Why is "sizeof(unsigned int) == sizeof(int)" refused in a cout?

Question

Why can I write:

bool a = sizeof(unsigned int) == sizeof(int);
cout << "(taille unsigned integer = integer) ? " << a;

But this:

cout << "(taille unsigned integer = integer) ? " << sizeof(unsigned int) == sizeof(int);

produces a compilation error?

Invalid operands to binary expression ('std::basic_ostream<char>::__ostream_type' (aka 'basic_ostream<char, std::char_traits<char>>') and 'unsigned long')

Does this answer your question? https://stackoverflow.com/q/20767745/4117728 — 463035818_is_not_an_ai, Sep 27 '22 at 06:41
On top of everything - the first version is a lot easier to read and understand. Pretty obvious that you want to print a bool. Not all that clear for the second version. — BoP, Sep 27 '22 at 08:00

score 4 · Accepted Answer · answered Sep 27 '22 at 06:47

This is an issue of operator precedence. The << operator has higher prcedence than ==, so your expression is parsed as

(cout << "(taille unsigned integer = integer) ? " << sizeof(unsigned int)) == (sizeof(int))

Since the ostream << operator overloads return the ostream they're called on, you're trying to compare a std::ostream to an int, and there is no such comparison.

score 3 · Answer 2 · edited Sep 27 '22 at 17:15

3

Due to operator precedence, it is interpreted as:

(cout << "? " << sizeof(unsigned int) ) == sizeof(int);

To solve that, add parentheses around ==:

cout << "? " << (sizeof(unsigned int) == sizeof(int));

edited Sep 27 '22 at 17:15

Remy Lebeau

555,201
31
458
770

answered Sep 27 '22 at 06:40

Quimby

17,735
4
35
55

Why is "sizeof(unsigned int) == sizeof(int)" refused in a cout?

2 Answers2