Sizeof vs Strlen

44
votes
#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[]) {
    char string[] = "october"; // 7 letters

    strcpy(string, "september"); // 9 letters

    printf("the size of %s is %d and the length is %d\n\n", string,
        sizeof(string), strlen(string));

    return 0;
}

Output:

$ ./a.out
the size of september is 8 and the length is 9

Is there something wrong with my syntax or what?

4
csizeofstrlen
You are writing past the end of the array string. This is undefined behavior. string can only hold 8 characters (7 for "october" and 1 for the null terminator). When you call strcpy, you are writing 10 characters to it (9 for "september" and 1 for the null terminator), which means you have gone past the end of the array and are overwriting the adjacent memory. - Marlon
Note that sizeof is calculated at compile time where as strlen is run time. - Naveen
@Naveen: Be aware that that is not necessarily true where VLAs are involved. - caf
@caf: Maybe I'm gonna feel clueless but..... What do you mean by VLAs? Very large arrays? lol - Cory Gross
I am saying that in the function int foo(int n) { int a[n]; the value sizeof a is not calculated at compile-time. - caf

4 Answers

41
votes

sizeof and strlen() do different things. In this case, your declaration

char string[] = "october";

is the same as 

char string[8] = "october";

so the compiler can tell that the size of string is 8. It does this at compilation time.

However, strlen() counts the number of characters in the string at run time. So, after you call strcpy(), string now contains "september". strlen() counts the characters and finds 9 of them. Note that you have not allocated enough space for string to hold "september". This is undefined behaviour.

2
votes

The Output is correct because

first statement string size was allocated by compiler that is 7+1 (October is 7 bytes & 1 byte for null terminator at compile time)

Second statement: you are copying September (9 bytes to 8 bytes string);

there for you got size of September as 8 bytes (still strlen() will not work for September it does not have null character)

1
votes

Your destination array is 8 bytes (length of "october" plus \0) and you want to put in 9 chars in that array.

man strcpy says: If the destination string of a strcpy() is not large enough, then anything might happen.

Please tell me what you really want to do, because this smells bad long way

-1
votes

You must eliminate buffer overflow problem in this example. One way to do this - is to use strncpy:

memset(string, 0, sizeof(string));
strncpy(string, "september", sizeof(string)-1);