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Why I cannot free a malloc'd string?

I have this code:

#define ABC "abc"

void main()
{
char *s = malloc(sizeof(char)*3);
printf("%p ", s);
s = ABC;
printf("%p ", s);
free(s);
}

This is the output:

0x8927008 0x8048574 Segmentation fault (core dumped)

As you can see, the address of string s changes after assignment (I think this is why free() gives segfault). Can anyone explain me why and how this happens? Thank you!

like image 756
antonio Avatar asked Mar 31 '26 15:03

antonio


1 Answers

The line

s = ABC;

changes s to point to a different string which may well be in read-only memory. Attempting to free such memory results in undefined behaviour. A crash is likely.

I think you wanted

strcpy(s, ABC);

instead. This would copy the char array "abc" into s. Note that this will cause a further bug - s is too short and doesn't have space for the nul terminator at the end of ABC. Change you allocation to 4 bytes to fix this

char *s = malloc(4);

or use

char *s = malloc(sizeof(ABC));

if ABC is the max length you want to store.

like image 165
simonc Avatar answered Apr 02 '26 10:04

simonc



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