Segmentation fault in my strcpy

Question

I'm writing my own strcpy due to the fact that the default one in string.h only accept a const char * as a source string to copy from. I'm trying this very basic prototype (yes, the return isn't meaningful, I'm just trying things):

int copyStrings(char * dest, char * source){
    int i=0;
    while(source[i]!='\0'){
        dest[i]=source[i];
        i++;
    }
    dest[i]='\0';
    return 0;
}

and it gives me SIGSEGV, Segmentation Fault error in gdb, at the line dest[i]=source[i], right at the first character. I'm pretty sure dest[i] isn't a string literal, so I should be able to modify it.

What am I doing wrong?

EDIT: here's the calling

int main(){
    char * str = (char*)malloc((int)sizeof(double));
    char * str2 = (char *)malloc((int)sizeof(int));
    str = "hello";
    str2 = "hey jude";
    copyStrings(str2, str);
    free(str);
    free(str2);
    return 0;
}

Answer 1

This is assigning a string literal to str2 - the very thing that you claim you aren't doing. This is actually the cause of your segfault.

str2 = "hey jude";

It also is causing a memory leak as prior to this, you malloc'd some memory and assigned it to str2 as well. But not enough memory to hold the string. Typically an int is 4 bytes and you need 9 bytes to store that string.

What you want to do is this, which allocates as many bytes as there are in the string, plus an extra one to store the \0 terminating character at the end.

str2 = malloc(strlen("hey jude")+1);
strcpy(str2,"hey jude");

or on some systems you can use POSIX function strdup() which effectively does the job of the above in one handy function call.

str2 = strdup("hey jude");