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Why does this work? Character pointer in C

#include <stdio.h>
#include <stdlib.h>
#include <iostream>

int main()
{
    char* s = (char*)malloc(sizeof(char) * 3); //I allocate memory for 3 chars
    s[0] = 'a';
    s[1] = 'b';
    s[2] = '\0';
    s[3] = 'd'; //This shouldn't work

    std::cout << s[3] << std::endl; //It prints out d, why?

    free(s);        
    return 0;
}

Why am I allowed to write s [3]?

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8 answers

Usually the malloc function allocates a certain minimum number of bytes, for example 16, even if you ask to select only one (or 3 as in your example) byte (s).

However, you should not rely on such a function function. An implementation is defined as malloc allocates memory. So your code has undefined behavior.

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This is called undefined behavior.

The compiler does not prevent you from committing illegal actions; however, compiled code may not do what you expect.

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