[英]Converting std::string::length to int
string :: length的返回類型為size_t,但似乎可以將其放入int中,而無需進行任何強制類型轉換。 在這種情況下,為什么要為int分配一個size_t?
int main() {
string line;
getline(cin, line);
cout << line << endl;
int i = line.size();
int j = line.length();
cout << i << " " << j << endl;
}
size_t
值正在縮小。 在c ++ 11中,您可以通過執行以下操作使此失敗並出現錯誤:
#include <string>
int main() {
std::string line;
int i{line.size()};
int j{line.length()};
}
產生的錯誤如下所示:
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>( )
size_t是32位整數。 轉到編譯器目錄並打開stdio.h文件。
有一個聲明,像這樣:
typedef unsigned int size_t;
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