Why should I declare a String as a fixed-size array

Question

I have this program:

#include<stdio.h>

void copy_string(char string1[], char string2[]){
    int counter=0;
    while(string1[counter]!='\0'){
        string2[counter] = string1[counter];
        counter++;
    }
    string2[counter] = '\0';
}

int main() {
   char* myString = "Hello there!";
   char* myStringCopy;
   copy_string(myString, myStringCopy);
   printf("%s", myStringCopy);
}

My question is, why isn't it working unless I declare myStringCopy as a fixed-size variable (char myStringCopy[12];)? Shouldn't it work if I add a \0 character after the copy as I'm doing?

The problem is that ther is no space allocated for `myStringCopy`. Its just an unitnitzialized pointer which points to no valid memery location, therefore you can not write to it. — Osiris, Aug 01 '18 at 14:35

Miguel · Answer 1 · 2018-08-01T15:30:53.660

4

It can work by doing char* myStringCopy as long as you allocate memory space for it.

for example

char* myStringCopy
myStringCopy = malloc(sizeof(char) * (strlen(myString)+1))

I might be mistaken about the +1 but I think it is like this.

edited Aug 01 '18 at 15:30

answered Aug 01 '18 at 14:37

Miguel

1,295
12
25

1

[Dont cast malloc return](https://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc). – H.S. Aug 01 '18 at 15:15

score 3 · Accepted Answer · answered Aug 01 '18 at 14:39

char myStringCopy[12]; tells the compiler to create an array of 12 char. When myStringCopy is passed to copy_string, this array is automatically converted to a pointer to its first element, so copy_string receives a pointer to the characters.

char *myStringCopy; tells the compiler to create a pointer to char. The compiler creates this pointer, including providing memory for it, but it does not set the value of the pointer. When this pointer is passed to copy_string, copy_string does not receive a valid value.

To make char *myStringCopy; work, you must allocate memory (which you can do with malloc). For example, you could use:

char *myStringCopy;
myStringCopy = malloc(13 * sizeof *myStringCopy);
if (myStringCopy == NULL)
{
    fprintf(stderr, "Error, the malloc did not work.\n");
    exit(EXIT_FAILURE);
}

Also, note that 12 is not enough. The string “Hello there!” contains 12 characters, but it also includes a terminating null character. You must provide space for the null character. char myStringCopy[12]; appeared to work, but copy_string was actually writing a thirteenth character beyond the array, damaging something else in your program.

@CarlosMorente: You can find the string length by counting the characters in the source string, up to and including the terminating null character. Also, if you want `copy_string` to do the allocation, then it must return the newly allocated address to the caller. — Eric Postpischil, Aug 01 '18 at 14:44

score 2 · Answer 3 · answered Aug 01 '18 at 14:37

2

The problem is that you don't have room for mystringCopy

You need to reserve space first:

   char* myString = "Hello there!";
   char* myStringCopy = malloc(strlen(myString) + 1);

answered Aug 01 '18 at 14:37

David Ranieri

39,972
7
52
94

score 1 · Answer 4 · answered Aug 01 '18 at 14:40

1

char* myStringCopy;

This is only pointer to char*. You must first allocate memory for myStringCopy, before start copy. When you declare it like this:

char myStringCopy[12];

compiler allocate enough memory in stack.

answered Aug 01 '18 at 14:40

Dnii Volna

31
3

i know, i only answer to question. and copy statement from user code. – Dnii Volna Aug 01 '18 at 14:43

Why should I declare a String as a fixed-size array

4 Answers4