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If the following program (myprog) is run from the command line as myprog 1 2 3 what would be the output? main(int argc, char *argv[]) { int i, j = 0; for (i = 0; i < argc ; i++) j = j + atoi ( argv[i]); printf ("%d", j); }

If the following program (myprog) is run from the command line as myprog 1 2 3 what would be the output? main(int argc, char *argv[]) { int i, j = 0; for (i = 0; i < argc ; i++) j = j + atoi ( argv[i]); printf ("%d", j); } Correct Answer 6

When atoi() tries to convert argv[0] to a number it cannot do so (argv[0] being a file name) and hence returns a zero.

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What is the output of this program ? #include
using namespace std;
int n(char, int);
int (*p) (char, int) = n;
int main()
{
(*p)('d', 9);
p(10, 9);
return 0;
}
int n(char c, int i)
{
cout << c << i;
return 0;
}
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