Really weird if statement behavior in C++

Question

Only the first statement is printed even though both if expressions should evaluate the same. Why?

    map <int, int> elem;
    vector <int> nums(2);
    int x = 1;
    elem[x]=-1;
    
    int val = elem[x];

    if(val>nums.size()/3) {cout << "First";} // as you can see, vals is -1 and nums.size() is 2
        
    if(-1>2/3) cout<< "Second";

If you are not asking for help with a compilation error, it's a good idea to make sure your example code compiles [without warnings](https://stackoverflow.com/questions/57842756/why-should-i-always-enable-compiler-warnings). — JaMiT, Sep 23 '20 at 03:56

score 3 · Accepted Answer · answered Sep 23 '20 at 01:41

3

They are not the same.

if(val > nums.size() / 3) {

The problem is that num.size() has type of size_t which is unsigned. When you compare signed to unsigned type, you get undefined behaviour.

If you really want to make it the same, then you need to do a cast:

if (val > static_cast<int>(nums.size()) / 3)

now it would be similar to your second code

if(-1 > 2 / 3)

both of which would result in false and you should not see any First or Second printing.

answered Sep 23 '20 at 01:41

artm

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1

Or: `if(-1 > static_cast(2) / 3)` – Remy Lebeau Sep 23 '20 at 01:45
2

This is not undefined behavior, see https://stackoverflow.com/questions/5416414/signed-unsigned-comparisons – zkoza Sep 23 '20 at 02:33

Really weird if statement behavior in C++

1 Answers1