is auto type detection only looking at one statement

user_724111 0
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For the two for loop below, are they equivalent? Basically does the auto look over "=0" assignment and see iter is compared with s.size() and so decides iter is of type decltype(s.size())?

string s;
for(auto iter=0; iter<s.size();iter++)

string s;
for(decltype(s.size()) iter=0; iter<s.size();iter++)
auto c++11 c++
3 Answers

and see iter is compared with s.size() and so decides iter is of type decltype(s.size())?

No. auto only deduces type from the initializer, given auto iter=0;, the type is deduced from 0 then it'll be int.

in the type specifier of a variable: auto x = expr;. The type is deduced from the initializer.

And you can specify the type with auto like:

auto iter = 0u;  // unsigned int
auto iter = 0ul; // unsigned long int
auto iter = 0uz; // std::size_t, since C++23

BTW in decltype(s.size()) iter=0;, the type would be deduced from s.size() based on the rule of decltype, it won't be influenced by the fact that iter is compared with s.size() later either.

user_33097900

Nope. Whenever a variable declaration has a "placeholder type" (auto, decltype(auto), const auto&, etc.), only that variable's initializer is used to deduce the actual type.

So in the first snippet, the type of iter is decltype(0) is int.

user_4596400

No, the loops are not equivalent. auto deduces the type only from the initializer. In your loop, that's 0 and 0 has the type int. So your first loop is

for(int iter=0; iter<s.size();iter++)

while the second loop is

for(conatiner_size_type iter=0; iter<s.size();iter++)

where conatiner_size_type is most likely a std::size_t

Depending on what you are doing, you may want to use a ranged based for loop like

for(auto character : s)
    // loop logic

and this will loop through the string and character will be the character at the current position in the string the loop is at.

user_43424980
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