C+中的_tmain()和main()有什么區別？如果我使用以下main()方法運行我的C+應用程序，一切正常：int main(int argc, char *argv[]) {
cout << "There are " << argc << " arguments:" << endl;
// Loop through each argument and print its number and value
for (int i=0; i<argc; i++)
cout << i << " " << argv[i] << endl;
return 0;}我得到了我所期望的，我的論點被打印出來了。但是，如果我使用_tmain：int _tmain(int argc, char *argv[]) {
cout << "There are " << argc << " arguments:" << endl;
// Loop through each argument and print its number and value
for (int i=0; i<argc; i++)
cout << i << " " << argv[i] << endl;
return 0;}它只顯示每個參數的第一個字符。造成這種情況的原因是什么？
3 回答
明月笑刀無情
TA貢獻1828條經驗 獲得超4個贊
int _tmain(int argc, _TCHAR *argv[])
int wmain(int argc, wchar_t *argv[])
int main(int argc, char *argv[])
int wmain(int argc, char *argv[])
#include <iostream>#include <tchar.h>#if defined(UNICODE)
#define _tcout std::wcout#else
#define _tcout std::cout#endifint _tmain(int argc, _TCHAR *argv[]) {
_tcout << _T("There are ") << argc << _T(" arguments:") << std::endl;
// Loop through each argument and print its number and value
for (int i=0; i<argc; i++)
_tcout << i << _T(" ") << argv[i] << std::endl;
return 0;}
幕布斯6054654
TA貢獻1876條經驗 獲得超7個贊
cout << _T( "There are " ) << argc << _T( " arguments:" ) << endl;
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