我对字符串的理解有什么问题，即在C编程中以'\\ 0'元素结尾的字符数组

Question

I am a total beginner and recently started studying strings in C programming.

I understand that we need to supply the '\\0' (null) character at the end of the string (to act as a end of string marker). So if my character array is

char string[]={'H','E','L','L','O','\0'};

This makes it a 6 element array.

So I was going through this simple example of copying one string to another string. Here is the code.

#include<stdio.h>

int main()
{

    char string1[80],string2[80];
    int i;

    printf("Enter a string \n");
    gets(string2);

    for(i=0; string2[i]!= '\0';i++)
    {
        string1[i]=string2[i];

    }
    string1[i]='\0';  /*here is my problem*/
    printf("The copied string is \n");
    printf("%s",string1);
    printf("\n");
    printf("The number of character are \t");
    printf("%d \n",i);

}

why isn't it string1[i+1]='\\0' ?? I mean, isn't by putting string1[i]='\\0' overwrite the last element that was just stored in the above for loop?

Answer 1

The code is correct, because of the way the for loop works:

for(A; B; C)
    CODE;

Is equivalent to:

A;
while (B)
{
    CODE;
    C;
}

(Except for the use of continue , that will jump to the increment expression, not the condition, as it would happen with a while ).

And since the loop ends when string2[i] != '\\0' , it is obvious that upon exiting, i is the index for the proper NUL byte. So after that:

string1[i] = '\0'; 

will write the \\0 at the same place as it is in string2 .

---

Usually for this kind of analysis it is helpful to think about preconditions and postconditions . That is, assuming there are no break and no goto , you are guaranteed that at the beginning of a for or while loop the condition is always true . And just after the end of a for or while loop the condition is always false .

Your particular code, illustrated with assert calls:

for(i=0; string2[i] != '\0'; i++)
{
    asssert(string2[i] != '\0');
    string1[i] = string2[i];
}
asssert(string2[i] == '\0');
string1[i] = '\0';

Answer 2

Looking at the code, it seems as if i is incremented as long as the character is not '\\0'. So, the last time it's incremented, it increments to a new position that hasn't had a character written to it yet.

So, at the end of the loop, there's a space to write the null character to.