String find function in C++ returns 0 if string passed as argument is empty

Question

When a empty string is passed to find function it returns 0.
If uninitialized string is passed, then also it returns 0.

#include <iostream>
#include <string>
using namespace std;

int main() {
    string str1="";
    string str2="This is a sample string";
    unsigned int loc = str2.find(str1);
    cout << "Loc : " << loc << endl;
    if(loc != string::npos)
    {
        cout << "Found" << endl;
    }
    else
    {
        cout << "Not found" << endl;
    }
    return 0;
}

Output :
Loc: 0
Found

My question is why does find return 0 instead of returning string::npos?

Answer 1

A needle l can be found in position i of haystack h if i + len(l) <= len(h) and for all 0 <= k < len(l) we have l[k] == h[i+k] . What you're dealing with is simply the vacuous truth of this definition: 'for all x ' automatically becomes true if there's no valid x to begin with.

Answer 2

As the docs on cppreference say. (Emphasis is mine.)

Finds the first substring equal to the given character sequence. Search begins at pos, ie the found substring must not begin in a position preceding pos.

1. Finds the first substring equal to str .

When is a string found at a position?

Formally, a substring str is said to be found at position xpos if all of the following is true:

• xpos >= pos
• xpos + str.size() <= size()
• for all positions n in str , Traits::eq(at(xpos+n), str.at(n))

An empty string fulfills these requirements at position 0.

The docs further say. (Emphasis is mine.)

In particular, this implies that

• a substring can be found only if pos <= size() - str.size()
• an empty substring is found at pos if and only if pos <= size()
• for a non-empty substring, if pos >= size() , the function always returns npos .

Answer 3

It found the exact match at position 0 (the first one searched), so that index is returned.

Quoting from cppreference , the conditions are:

Formally, a substring str is said to be found at position xpos if all of the following is true:

1. xpos >= pos
2. xpos + str.size() <= size()
3. for all positions n in str , Traits::eq(at(xpos+n), str.at(n))

Trying to match substring from position 0, we get that

1. is met, because 0 (our search position) >= 0 (start parameter)
2. is met, because 0 + 0 <= size()
3. is met, because there is no character in the needle that doesn't match the corresponding character in haystack - all of them match.