In the following declaration of main, "int main(int argc,
char *argv[])", to what does argv[0] usually correspond?
1) The first argument passed into the program
2) The program name
3) You can't define main like that
Answers were Sorted based on User's Feedback
Answer / sanish joseph
No guys the ans s 2
it prints d program name:
arguments counts only from 1 nd nt 4m 0
| Is This Answer Correct ? | 15 Yes | 0 No |
Answer / vadivel
If the value of argc is greater than 0, the array members
argv[0] through argv[argc-1] inclusive shall contain
pointers to strings, which will be the command line
arguments. argv[0] will contain the program name, argv[1]
the first command line arg, argv[1] the second and so on.
Finally, argv[argc] is guaranteed to the a NULL pointer,
which can be useful when looping through the array.
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / sumeet sharma
argv[o] the array of strings will point to the name of the
program.
argv[1] will point to the first argument and so on.
| Is This Answer Correct ? | 1 Yes | 0 No |
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