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Home/ Questions/Q 6078843
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T10:52:45+00:00 2026-05-23T10:52:45+00:00

I’m trying without success to copy a char array to another one. I have

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I’m trying without success to copy a char array to another one. I have tried memcpy copying direcly the address from one to another, like this:

void include(int id, char name[16]) {
int i;

  for (i = 0; i < SZ; i++) {
      if (a[i].id == 0) {
          a[i].id = id;
          memcpy(&a[i].name, &name, strlen(name)+1);
          return;
      }
  }
}

But obviously works only inside of this function. I have tried also like this: http://www.cplusplus.com/reference/clibrary/cstring/memcpy/ but it didn’t work.
Can someone help me?

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  1. Editorial Team
    Editorial Team
    2026-05-23T10:52:45+00:00Added an answer on May 23, 2026 at 10:52 am

    Drop the & from &name and it should work. Your function declaration is misleading; it’s actually equivalent to:

    void include(int id, char *name)
    

    The compiler pretends that the array
    parameter was declared as a pointer

    If name would be an array, name == &name. But name is a pointer so name != &name.

    The C FAQ has some questions that might help:

    • Array parameters
    • Array parameter size
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