c assignment makes pointer without a cast

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[SOLVED] C - assigment makes integer from pointer without a cast warning

Thread: [solved] c - assigment makes integer from pointer without a cast warning, thread tools.

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I know it's a common error, and I've tried googling it, looked at a bunch of answers, but I still don't really get what to do in this situation.... Here's the relevant code: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char path[51]; const char* home = getenv( "HOME" ); strcpy( path, argv[1] ); path[1] = home; return 0; } -- there is more code in the blank lines, but the issue's not there (I'm fairly sure), so didn't see the point in writing out 100 odd lines of code. I've tried some stuff like trying to make a pointer to path[1], and make that = home, but haven't managed to make that work (although maybe that's just me doing it wrong as opposed to wrong idea?) Thanks in advance for any help

Re: C - assigment makes integer from pointer without a cast warning

path[1] is the second element of a char array, so it's a char. home is a char *, i.e. a pointer to a char. You get the warning because you try to assign a char* to a char. Note also the potential to overflow your buffer if the content of the argv[1] argument is very long. It's usually better to use strncpy.

Last edited by r-senior; March 10th, 2013 at 03:03 PM . Reason: argv[1] would overflow, not HOME. Corrected to avoid confusion.

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You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that.

Last edited by Christmas; March 10th, 2013 at 09:51 PM .

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Originally Posted by Christmas You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that. Excellent point. I've basically fixed my problem by reading up on pointers again (haven't done any C for a little while, so forgot some stuff), and doing: Code: path[1] = *home; the code doesn't moan at me when I compile it, and it runs okay (for paths which aren't close to 51 at least), but after reading what you read, I just wrote a quick program and found out that getenv("HOME") is 10 characters long, not 1 like I seem to have assumed, so I'll modify my code to fix that.

Yes, getenv will return the path to your home dir, for example /home/user, but path[1] = *home will still assign the first character of home to path[1] (which would be '/').

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Converting Pointers to Integers: Avoiding Cast Errors & Mastering the Process

In this guide, we will cover how to convert pointers to integers and vice versa without running into any cast errors. We will also walk you through the step-by-step process of mastering pointer and integer conversions in C/C++.

Table of Contents

• Why Convert Pointers to Integers
• Understanding uintptr_t
• Step-by-Step Guide
• Converting Pointers to Integers
• Converting Integers to Pointers

Why Convert Pointers to Integers? {#why-convert-pointers-to-integers}

There are several use cases where you might need to convert pointers to integers and vice versa. Some common reasons include:

• Manipulating memory addresses for low-level programming.
• Serializing and deserializing data.
• Storing pointers in a generic data structure.
• Debugging and logging purposes.

However, when converting pointers to integers, it is crucial to avoid any errors that may arise from incorrect casting.

Understanding uintptr_t {#understanding-uintptr_t}

To safely convert pointers to integers, it is essential to use the uintptr_t data type. This is an unsigned integer type that is large enough to store the value of a pointer. It is available in the <stdint.h> header in C and the <cstdint> header in C++.

Using uintptr_t , you can safely cast a pointer to an integer and back to a pointer without losing any information. This ensures that the process is safe, fast, and efficient.

Step-by-Step Guide {#step-by-step-guide}

Converting pointers to integers {#converting-pointers-to-integers}.

To convert a pointer to an integer, follow these steps:

• Include the <stdint.h> header (C) or the <cstdint> header (C++) in your program.
• Cast your pointer to uintptr_t .

Converting Integers to Pointers {#converting-integers-to-pointers}

To convert an integer to a pointer, follow these steps:

• Cast your integer to the required pointer type using a double cast.

FAQs {#faqs}

Why can't i just use a regular int or unsigned int to store pointers {#regular-int}.

While it may work on some platforms where the size of an int is equal to the size of a pointer, it is not guaranteed to be portable across different systems. Using uintptr_t ensures your code remains portable and safe.

Are there performance implications when using uintptr_t ? {#performance}

The performance impact of using uintptr_t is minimal. Most modern compilers can optimize the casting operations, resulting in little to no overhead.

When should I use intptr_t instead of uintptr_t ? {#intptr_t}

intptr_t is a signed integer type that can hold a pointer value. It is useful when you need to perform arithmetic operations on pointers that may result in negative values. However, in most cases, uintptr_t is recommended.

Is it safe to perform arithmetic operations on integers representing pointers? {#pointer-arithmetic}

Performing arithmetic operations on integers representing pointers can lead to undefined behavior if the resulting integer doesn't correspond to a valid memory address. It is generally safer to perform arithmetic operations on pointers directly.

How do I avoid losing information when casting pointers to integers? {#avoid-losing-information}

By using uintptr_t , you ensure that the integer is large enough to store the value of a pointer without losing any information. Make sure always to use uintptr_t when converting pointers to integers.

Related Links

• C++ Reference: uintptr_t
• C Reference: uintptr_t
• Understanding Pointers in C and C++

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Error: assignment makes pointer from integer without a cast

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Thread: Error: assignment makes pointer from integer without a cast

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the following program returns error code: distinct.c:18: error: assignment makes pointer from integer without a cast. Code: #include <stdio.h>#include <stdlib.h> #define MAX_LENGTH 128 #define MAX_LINES 1024 int main(int argc, char *argv[]) { int i, j; int countFgets = 0; int count = 0; int numberofDistinctLines = atoi(argv[1]); char* string[MAX_LINES]; for(i = 0; (i < MAX_LINES) && fgets(string[i], MAX_LENGTH, stdin) != NULL; ++i) { ++countFgets; j = 0; count = 0; for (j = 0; string[j] != '\0'; ++j){ if ((string[i] =! string[j])){ ++count; if (count == numberofDistinctLines){ printf("%c distinct lines seen after %d lines read.\n", numberofDistinctLines, countFgets); return 0; } } } } return 0; } Please help!!!

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On line 19, you are comparing two strings with the operator !=. Wrong. You should use strcmp. Welcome to the forum!

Last edited by std10093; 05-11-2013 at 07:25 AM .

Code - functions and small libraries I use It’s 2014 and I still use printf() for debugging. "Programs must be written for people to read, and only incidentally for machines to execute. " —Harold Abelson

Originally Posted by std10093 On line 19, you are comparing two strings with the operator != Or rather, that was probably what you were trying to do. In reality, you wrote this: Code: if ((string[i] =! string[j])){ which can be written as: Code: if (string[i] = !string[j]){ which is equivalent to: Code: string[i] = !string[j]; if (string[i]){ The result of !string[j] is an int, either 0 or 1. You then convert this int to a pointer, without a cast, in order to assign it to string[i]. Also, this is not good practice: Code: for (j = 0; string[j] != '\0'; ++j){ string is an array of pointers to char. Therefore, string[j] is a pointer to char. You compared string[j] to '\0', but '\0' is a character literal (i.e., of type int). If you want to check if string[j] is not a null pointer, either write: Code: for (j = 0; string[j]; ++j){ or more explicitly: Code: for (j = 0; string[j] != NULL; ++j){

Originally Posted by Bjarne Stroustrup (2000-10-14) I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool. Look up a C++ Reference and learn How To Ask Questions The Smart Way

Thanks for the response. But i am still new to C and you solution didnt work (i fixed the NULL) but i don't know the format for strcmp or the second solution. Could you give a little more detail? For more information i have to created a C program that: A C program that when given a single argument n reads lines from standard input until n different lines

i don't know the format for strcmp It is part of your job to find these things out - and you dont need to ask questions in forums to do that - there are masses of documentation you can easily find

Thought for the day: "Are you sure your sanity chip is fully screwed in sir?" (Kryten) FLTK: "The most fun you can have with your clothes on." Stroustrup: "If I had thought of it and had some marketing sense every computer and just about any gadget would have had a little 'C++ Inside' sticker on it'"

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Error: assignment makes integer from pointer without a cast, assignment makes pointer from integer without a cast, assignment makes pointer from integer without a cast, warning: assignment makes integer from pointer without a cast.

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Makes Pointer From Integer Without a Cast: Fix It Now!

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Our research team has ensured that this will be your best resource on these error messages, and you won’t have to read another article about it again. With that said, launch your code that’s showing the error messages, and let’s fix it for you.

JUMP TO TOPIC

– You Assigned an Integer to a Pointer

– you want to convert an integer to a pointer, – you passed a variable name to the “printf()” function, – you used “struct _io_file” in a wrong way, – you copied a string to an invalid location, – you’re setting a pointer to a different type, – use equal data types during assignment, – ensure the pointer and integer have the same sizes, – pass a “format string” to the “printf()” function, – use a function that returns a pointer to “struct _io_file”, – copy the string to character array or character pointer, – assign pointers of compatible types, why your code made a pointer from an integer without a cast.

Your code made a pointer from an integer without a cast because you assigned an integer to a pointer or you want to convert an integer to a pointer. Other causes include the passing of a variable name to the “printf()” function and using “struct _io_file in the wrong way.

Finally, the following are also possible causes:

• You copied a string to an invalid location
• You’re setting a pointer to a different type

If you assign an integer to a pointer, it will lead to the “ assignment makes pointer from integer without a cast c programming ” error. For example, in the following, the “num_array” variable is an unsigned integer type while “tmp_array” is a pointer.

Later, an error will occur when the “for” loop tries to copy the values of the “num_array” to the “tmp_array” pointer using a variable assignment.

For example, in the following, the “theta” variable is an integer while “ptr_theta” is a pointer. Both have different sizes, and you can’t convert the integer to a pointer.

If you pass a variable name to the “printf()” function, that’s when the compiler will throw the “ passing argument 1 of ‘printf’ makes pointer from integer without a cast ” error.

For example, in the following, “num_val” is an integer variable, and the code is trying to print it using the “printf()” function.

When you assign an integer to a pointer of “struct _io_file”, that’s when you’ll get the “ assignment to file aka struct _io_file from int makes pointer from integer without a cast ” error. For example, in the following code, “FILE” is a “typedef” for “struct _io_file”. This makes it a pointer to “struct _io_file”.

Later, the code assigned it to the integer “alpha,” and this will lead to an error stated below:

Now, in the following, the code is trying to copy a string (“source”) into an integer (“destination”), and this leads to an error because it’s not a valid operation:

When your code sets a pointer to a different type, your compiler will show the “ incompatible pointer type ” error. In the following code, “pacifier” is an integer pointer, while “qwerty” is a character pointer.

Both are incompatible, and your C compiler will not allow this or anything similar in your code.

How To Stop a Pointer Creation From an Integer Without a Cast

You can stop a pointer creation from an integer without a cast if you use equal data types during the assignment or ensure the integer and pointer have the same sizes. What’s more, you can pass a “format string” to “printf()” and use a function that returns a pointer to “struct _io_file”.

• Copy the string to a character array or character pointer
• Assign pointers of compatible types

During a variable assignment, use variables of equal data types. In our first example, we assigned a pointer (uint8_t *tmp_array) to an unsigned integer (uint8_t num_array) and it led to a compile-time error.

Now, the following is the revised code, and we’ve changed “tmp_array” to a real array. This will prevent the “ gcc warning: assignment makes pointer from integer without a cast ” error during compilation.

Now, the fix is to use “intptr_t” from the “stdint.h” header file because it guarantees that the pointer and integer will have the same sizes. We’ve used it in the following code, and you can compile it without an error.

To fix the “pointer to integer without a cast” in the “printf()” function, pass a “format string” as its first argument. How you write the “format string” depends on the variable that you’ll print. In the following updated code, “num_val” is an integer, so the “format string” is “%d”.

When you’re using “FILE” from the “stdlib.h” header file, you can prevent any “pointer from integer” error if you use a function that returns a pointer to “struct _io_file”.

An example of such a function is “fopen()” which allows you to open a file in C programming. Now, the following is a rewrite of the code that causes the pointer error in “struct _io_file”. This time, we use “fopen()” as a pointer to “FILE”.

Now, in the following code, we’ve changed “destination” from an integer to a “character array”. This means “strcpy()” can copy a string into this array without an error.

Meanwhile, the following is another version that turns the “destination” into a “character pointer”. With this, you’ll need to allocate memory using the “malloc()” function from the “stdlib.h” header file.

When you’re assigning pointers, ensure that they’re compatible by changing their data type . The following is the updated code for the “charlie” example , and “qwerty” is now an integer.

This article explained why your code made a pointer without a cast and how you can fix it. The following is a summary of what we talked about:

• An attempt to convert an integer to a pointer will lead to the “makes pointer from integer without a cast wint conversion” error.
• To prevent an “incompatible pointer type” error, don’t set a pointer to a different type.
• If you’re using the “printf()” function, and you want to prevent any “pointer to integer” error, always pass a “format specifier”.

At this stage, you’ll be confident that you can work with integers and pointers in C programming without an error. Save our article, and share it with your developer communities to help them get rid of this error as well.

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The routine below draws an image on a tft-screen, and although the whole programm works fine and the image is displayed ok, I'm unable to resolve this 'assignment' warning at pS=pS0+n; .

I've tried several casts, but as yet not the right one (I'm not so well versed in casting).

Can anyone help with the solution?

Definition of the image:

Drawing the image:

The actual routine:

pS0 should have the same type as pS (const char *), and definitely not unsigned int.

TNKernel-PIC32, an open-source real-time kernel for the PIC32

I don't understand why you have ps0 defined as...

If you define it as the correct type, you can avoid all the ugly and unecessary casting:

Edit: andersm beat me to it.

It also seems a lot of your variables are incorrectly sized.  rgb8 could easily be unsigned char, for example.

You are trying to convert an pointer to an integer without cast and it is simply telling you that the correct type is what pS type is that is the thing that line is trying to set after all

That all said it would be a hell of a lot easier to make PS0 a const char* and simply tell it

Thank you all for the replies. I think this will help me resolve it. And the the other tips too!