What is the output of the following code:
int v()
{
int m=0;
return m++;
}
int main()
{
cout<<v();
}
1) 1
2) 0
3) Code cannot compile
4)
5)
by 
Anonymous
ANSWERS: 5
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46
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The answer is 2) 0.
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If you have the appropriate headers it will output 0.
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It prints nothing because there's no endl or other flush on the cout ;-) But this is a simple question about post increment. It boils down to whether return on a post incremented variable returns before or after the increment. The answer is that m is evaluated for the expression before the increment operator applies. Since m is then a static out of scope it is irrelevant, however, if m was a global or a call by ref it would be incremented and the pre increment value returned. So a slightly more interesting example is: int v(int &m) { return m++; } int main() { int m = 0; cout<<v(m)<<m; }
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m++ means read then increment ++m increment then read. Thus the result is zero.
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