How to make it not share the same memory

Question

#include <stdio.h>
typedef struct TESTCASE{
    char *before; 
}ts;
int main(void) {
     ts t[2] = {{"abcd"}, 
                {"abcd"}};
     t[0].before[0] = t[0].before[2] = t[0].before[3] = 'b';
     printf("%s - %s\n", t[0].before, t[1].before);
     return 0;
}

the output is

bbbb - bbbb

I compile with gcc in Cygwin

cc -g test.c -o test

my question is, with what compile option, I can reap a result of bbbb - abcd?

This question appears to be off-topic because it is a duplicate (but finding the duplicate is hard) and because it is unlikely to help other people in the future. — Jonathan Leffler, Mar 03 '14 at 03:05

score 5 · Answer 1 · edited May 23 '17 at 12:12

5

You are not supposed to write the strings, they are "immutable", writing to it results in undefined behaviour.

Because of that, the compiler is free to use the same location for both strings.

Hint: strdup() - what does it do in C?

edited May 23 '17 at 12:12

Community

1
1

answered Sep 20 '12 at 13:20

Karoly Horvath

94,607
11
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score 0 · Answer 2 · answered Sep 20 '12 at 13:36

0

t[0].before[3] = 'b'; will do a segmentation fault on some systems. You can't write into a constant string.

#include <stdio.h>
typedef struct TESTCASE{
  char before[5];
}ts;
int main(void) {
  ts t[2] = {{ {'a','b','c','d',0} },
             { {'a','b','c','d',0} }};
  t[0].before[0] = t[0].before[2] = t[0].before[3] = 'b';
  printf("%s - %s\n", t[0].before, t[1].before);
  return 0;
}

answered Sep 20 '12 at 13:36

Nahuel Fouilleul

18,726
2
31
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Linux and AIX (with XLC) being two such systems that will seg fault. – display101 Sep 20 '12 at 14:06

How to make it not share the same memory

2 Answers2