Command Line Arguments
- What does argc and argv indicate in command-line arguments?
(Assuming: int main(int argc, char *argv[]) )
-
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The name of the variable argc stands for "argument count"; argc contains the number of arguments passed to the program. The name of the variable argv stands for "argument vector". A vector is a one-dimensional array, and argv is a one-dimensional array of strings. Each string is one of the arguments that was passed to the program.
Correct Option: C
The name of the variable argc stands for "argument count"; argc contains the number of arguments passed to the program. The name of the variable argv stands for "argument vector". A vector is a one-dimensional array, and argv is a one-dimensional array of strings. Each string is one of the arguments that was passed to the program.
For example, the command line
gcc -o myprog myprog.c
would result in the following values internal to GCC:
argc
4
argv[0]
gcc
argv[1]
-o
argv[2]
myprog
argv[3]
myprog.c
- What will be the output of the following C code (run without any command line arguments)?
#include <stdio.h>
int main(int argc, char *argv[])
{
while (argv != NULL)
printf("%s\n", *(argv++));
return 0;
}
-
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NA
Correct Option: D
Segmentation fault/code crash
- What will be the output of the following C code (run without any command line arguments)?
#include <stdio.h>
int main(int argc, char *argv[])
{
while (*argv != NULL)
printf("%s\n", *(argv++));
return 0;
}
-
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NA
Correct Option: A
Executable file name
- What will be the output of the following C code (run without any command line arguments)?
#include <stdio.h>
int main(int argc, char *argv[])
{
while (*argv++ != NULL)
printf("%s\n", *argv);
return 0;
}
-
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NA
Correct Option: B
Segmentation fault/code crash
- What will be the output of the following C code (run without any command line arguments)?
#include <stdio.h>
int main(int argc, char *argv[])
{
printf("%s\n", argv[argc]);
return 0;
}
-
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NA
Correct Option: D
Segmentation fault/code crash