char pointers and arrays

This is a discussion on char pointers and arrays within the C Programming forums, part of the General Programming Boards category; while working on a simple little utility I came across something I don't quite understand why is it that Code: ...

  1. #1
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    char pointers and arrays

    while working on a simple little utility I came across something I don't quite understand
    why is it that
    Code:
     char str[] = "first,second,third";
    works for strtok()
    but
    Code:
      char *str = "first,second,third";
    ends in a segfault that even though I trace the stack I cant figure out why for the life of me

    Now it's been a while since I was in class for C, since I've been programming professionally for years but to the best of my knowledge
    char *blah = "this is a const char *";
    and char blah[] = "this is a const char*";
    were the same exact thing.

    The examples were compiled on a Linux x86 system
    anyway, here's the sample codes
    Code:
    #include <stdio.h>
    #include <string.h>
    
    int main(int argc, char *argv[])
    {
    	char str[] = "first,second,third";
    	char *result = NULL;
    	char *ptr = str;
    	printf("%s\n",str);
    	result=strtok(ptr,",");
    	while(result)
    	{
    		printf("%s\n",result);
    		result = strtok(NULL,",");
    	}
    	return 0;
    }
    the above WORKS
    Code:
    #include <stdio.h>
    #include <string.h>
    
    int main(int argc, char *argv[])
    {
    	char *str = "first,second,third";
    	char *result = NULL;
    	char *ptr = str;
    	printf("%s\n",str);
    	result=strtok(ptr,",");
    	while(result)
    	{
    		printf("%s\n",result);
    		result = strtok(NULL,",");
    	}
    	return 0;
    }
    the above results in a segmentation fault

  2. #2
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    1. You cannot modify string literal like "helo" it results in undefined behaviour.
    e.g char *p = "crash";
    *p = 'C'; // undefined behaviour!
    2. strtok() modifies the string. RTFM for more details.

  3. #3
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    while looking at the assemnly I learned that the way the compiler arranges the two char arrays
    for char *
    the string is set in static memory (i.e. .LC0 .string "blahblah")
    where as
    char []
    pushes the array onto the stack and copies the characters word (4bytes) at a time

    Code:
    char *
    .LC0:
    	.string	"first,second,third"
    ...
    	movl	$.LC0, 20(%esp)
    Code:
    char []
    	movl	$1936877926, 21(%esp)
    	movl	$1702046836, 25(%esp)
    	movl	$1684959075, 29(%esp)
    	movl	$1768453164, 33(%esp)
    	movw	$25714, 37(%esp)
    so the answer that I came up with in a simplified manner is that with a char *blah = "words" the characters are static and cannot be manipulated while char [] can be manipulated similar to the old const char * vs char *

    I cant believe it took me to disassemble the program to figure out the obvious.

  4. #4
    C++まいる!Cをこわせ! Elysia's Avatar
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    Posts
    22,583
    Save yourself trouble, declare all char* pointers that hold string literals const. Then you will get a compile error instead of a runtime error next time.
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.

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