Converting std::string::length to int

Question

string::length has the return type of size_t, but it seems to able to be put into an int without any casting or anything. Why can I assign a size_t to an int in this case?

int main() {
     string line;
     getline(cin, line);
     cout << line << endl;
     int i = line.size();
     int j = line.length();
     cout << i << " " << j << endl;
}

A good compiler should warn here, especially if the size type is larger than int. — Neil Kirk, Oct 18 '14 at 16:09
Remember, an `int` implies the value can go negative. I have yet to see negative string lengths. Better to use `unsigned int`. — Thomas Matthews, Oct 18 '14 at 17:17

score 3 · Accepted Answer · answered Oct 18 '14 at 16:05

3

The size_t values are being narrowed. In c++11, you could make this fail with an error by doing:

#include <string>

int main() {
    std::string line;
    int i{line.size()};
    int j{line.length()};
}

The errors produced look like:

gh.cc:5:11: error: non-constant-expression cannot be narrowed from type 'size_type' (aka 'unsigned long') to 'int' in initializer list [-Wc++11-narrowing]
    int i{line.size()};
          ^~~~~~~~~~~
gh.cc:5:11: note: override this message by inserting an explicit cast
    int i{line.size()};
          ^~~~~~~~~~~
          static_cast<int>( )
gh.cc:6:11: error: non-constant-expression cannot be narrowed from type 'size_type' (aka 'unsigned long') to 'int' in initializer list [-Wc++11-narrowing]
    int j{line.length()};
          ^~~~~~~~~~~~~
gh.cc:6:11: note: override this message by inserting an explicit cast
    int j{line.length()};
          ^~~~~~~~~~~~~
          static_cast<int>( )

answered Oct 18 '14 at 16:05

Bill Lynch

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Why is there no error for what was in my original post? They're essentially the same thing. – Matt Oct 18 '14 at 16:16
I don't get it. Why would you write code that you *know* it won't compile? – jrok Oct 18 '14 at 16:22
I'm just curious why one works and the other doesn't – Matt Oct 18 '14 at 16:34
@Math2: Let's look at forms 1 and 2 on this page: http://en.cppreference.com/w/cpp/language/direct_initialization Note that form 2 does not allow narrowing. – Bill Lynch Oct 18 '14 at 16:36
Thanks Bill. Seems to be a lot more rules in C++ than there is in Java – Matt Oct 18 '14 at 16:38
@Math2: Oh absolutely! – Lightness Races in Orbit Oct 18 '14 at 18:21

score 0 · Answer 2 · answered Oct 18 '14 at 16:48

0

size_t is a 32 bit integer. Go to your compiler directory and open the stdio.h file.

There is a declaration, something like this:

typedef unsigned int size_t;

answered Oct 18 '14 at 16:48

Martin Wantke

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2

This is an implementation detail. – John Dibling Oct 18 '14 at 17:20

Converting std::string::length to int

2 Answers2

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