复制字符串而不使用strcpy

Question

#include <stdio.h>
#include <stdlib.h>

int main() {
    char *str, *temp;
    str = malloc(sizeof(char) * 100);
    fgets(str, 100, stdin);
    temp = str;
    printf("%s", str);
    return 0;
}

Is this valid code to copy one string to another without using strcpy() function?

Answer 1

You are merely copying the starting address of str to temp . This means that any changes to temp will be reflected in str as well, since they point to the same memory. It does not truly emulate strcpy(dest, src) , which creates a separate copy of the null-terminated string pointed to by src starting at the memory location pointed to by dest .

So, to answer your question as asked: no.

If your intent is to avoid the O(n) running time of strcpy , that's also something you can't really do.

If you're required by a programming assignment or exercise to create code functionally equivalent to strcpy , here is a high-level description of the algorithm it uses:

1. Copy the character in *source to *destination
2. If the character that was just copied was the null terminator, exit.
3. Otherwise, increment the pointer to source , and increment the pointer to destination .
4. Go to step 1.