snprintf and sprintf explanation

Can someone explain the output of this simple program?

#include <stdio.h>

int main(int argc, char *argv[])
{
    char charArray[1024] = "";
    char charArrayAgain[1024] = ""开发者_如何学编程;;
    int number;

    number = 2;

    sprintf(charArray, "%d", number);

    printf("charArray : %s\n", charArray);

    snprintf(charArrayAgain, 1, "%d", number);
    printf("charArrayAgain : %s\n", charArrayAgain);

    return 0;
}

The output is:

./a.out 
charArray : 2
charArrayAgain : // Why isn't there a 2 here?

---

Because snprintf needs space for the \0 terminator of the string. So if you tell it the buffer is 1 byte long, then there's no space for the '2'.

Try with snprintf(charArrayAgain, 2, "%d", number);

---

snprintf(charArrayAgain, 1, "%d", number);
//                       ^

You're specifying your maximum buffer size to be one byte. However, to store a single digit in a string, you must have two bytes (one for the digit, and one for the null terminator.)

---

You've told snprintf to only print a single character into the array, which is not enough to hold the string-converted number (that's one character) and the string terminator \0, which is a second character, so snprintf is not able to store the string into the buffer you've given it.

---

The second argument to snprintf is the maximum number of bytes to be written to the array (charArrayAgain). It includes the terminating '\0', so with size of 1 it's not going to write an empty string.

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Check the return value from the snprintf() it will probably be 2.