CODEWITHSUNDEEP
clinux
Nikhil
Nikhil

Reputation: 586

memcpy function in c

Why this below code gives segmentation fault?

int main()
{
        char *t = "Working on RedHat Linux";
        char *s;

        s = malloc (8000 * sizeof(char));
        memcpy(s,t,7000);  
        printf("s = %s\nt = %s\n",s,t);
        free(s);
}

I have allocated 8000bytes for 's'. And copying only 't' to s untill 7000bytes. Though I have allocated 8000 bytes for 's', why its giving segmentation fault?

Upvotes: 1

Views: 2940

Answers (4)

EFraim
EFraim

Reputation: 13058

The segmentation fault is because t points to a region smaller than 7000 bytes. You are probably trying to read into an area when no readable page is mapped (after the end of string literal "Working on RedHat Linux"). You should limit your memcpy to sizeof("Working on RedHat Linux") bytes.

Upvotes: 11

ouah
ouah

Reputation: 145899

Use:

memcpy(s, t, strlen(t) + 1);

to avoid memcpy reading past the string literal array.

The C standard says regarding string functions (memcpy is a string.h function) (C99, 7.21.1p1).

"If an array is accessed beyond the end of an object, the behavior is undefined."

Upvotes: 2

Sumit Jain
Sumit Jain

Reputation: 1528

t points to a String buffer of length 24, but in memcpy you are trying to copy more than that (7000) which do not exist.

You are trying to access memory beyond what is allocated. So it is giving a segmentation fault

Upvotes: 1

Employed Russian
Employed Russian

Reputation: 213957

Your program exhibits undefined behavior: for memcpy() to work, both the source and the destination must be addressable for the number of bytes you've specified.

You've satisfied the destination, but not the source part.

Also, you can remove sizeof(char) as it is defined by the standard to always be 1.

Upvotes: 4

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