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

Q: 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); }

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.

Question

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); }

Accepted Answer

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.

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); }

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