why the result of strlen() out of my expection?-CodePudding

The function strlen() counts the number of characters in a string up to NUL and it doesn't contain NUL.In ASCII,NUl is equal to '\0'.

#include<stdio.h> 
#include<string.h>
int main(void)
{
    char a[]="abc";
    char b[]="abcd'\0'def";
    printf("%d\n%d",strlen(a),strlen(b));
    return 0;
}

The result is 3 and 5. THe second result is in contradiction with the first result. Thus ,I try to find how to implement strlen().

int strlen(char a[])
{
    int i;
    for(i=0;a[i]!='\0';i  );
    return i;
}

Based on this code ,I can understand the first result,but really can't understand the second one. Why is the sceond result not 4 but 5?Thanks in advance.

CodePudding user response：

You are getting 5 because you have wrapped the NUL character in single quotes, the value 5 is the length of the string "abcd`".

If you change the second example to "abcd\0ef" (no single quotes), you get a value of 4.

CodePudding user response：

char b[]="abcd'\0'def";

the array elements are

[a][b][c][d]['][0]['][d][e][f][0]

so the lenthth of the string before first [0] is 5 as it contains ' character as well.

CodePudding user response：

In ASCII, NUL is equal to '\0'.

NUL, and the null character, are equal to '\0', that is correct. But what you are looking here is a character literal (one character, \0, enclosed in single quotes). Within a string literal, those single quotes are not needed, and will indeed be interpreted as characters of their own.

So this...

char a[]="abc";
char b[]="abcd'\0'def";

...is equivalent to:

char a[]= { 'a', 'b', 'c', '\0' };
char b[]= { 'a', 'b', 'c', 'd', '\'', '\0', '\'', 'd', 'e', 'f', '\0' };

Your intention for b was this:

char b[]="abcd\0def";