请教关于cout和printf的一个问题~
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两段代码:
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#include <iostream> #include <windows.h> using namespace std; int main(int argc, char* argv[]) { int nSelect=::MessageBox(NULL,"Hello,Windows XP","Greetings",MB_OKCANCEL); if(nSelect==IDOK) cout<<"用户选择了“确定”按钮<<endl; else cout<<"用户选择了“取消”按钮<<endl; return 0; }
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#include "stdafx.h" #include <windows.h> int main(int argc, char* argv[]) { int nSelect=::MessageBox(NULL,"Hello,Windows XP","Greetings",MB_OKCANCEL); if(nSelect==IDOK) printf("用户选择了“确定”按钮\n"); else printf("用户选择了“取消”按钮\n"); return 0; }
VC++6.0 编译,第一段代码出错,错误如下:
error C2001: newline in constant error C2143: syntax error : missing ';' before 'else' error C2001: newline in constant error C2143: syntax error : missing ';' before 'return'
而第二段代码却没问题,正常运行
请问这是为什么?
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[ 本帖最后由 zjjne 于 2012-7-8 13:40 编辑 ]