What happens if you try to compile and run this program?
#include
int main (int argc, char *argv[]) {
int i = 0;
printf ("%s", argv[i]);
return 0;
}
Choose the right answer:
Options:
A.
The program outputs an unpredictable string, or execution fails
B.
The program outputs a predictable non-empty string
C.
Execution fails
D.
The program outputs an empty string
E.
Compilation fails
Answer:
B
Explanation:
Explanation:
The program is a valid C program that can be compiled and run without errors. The program uses the argc and argv parameters of the main function, which are used to pass command-line arguments to the program. The argc parameter is an integer that stores the number of arguments, including the name of the program itself. The argv parameter is an array of strings that contains the arguments. The first element of the array, argv[0], is always the name of the program. The program declares an integer variable i and assigns it the value of 0. Then it prints the value of argv[i] as a string using the %s format specifier. Since i is 0, this is equivalent to printing argv[0], which is the name of the program. Therefore, the program outputs a predictable non-empty string, which is the name of the program. The exact name of the program may vary depending on how it is compiled and executed, but it will not be an empty string, an unpredictable string, or cause an execution failure. References = Command Line Arguments in C - GeeksforGeeks, Program Arguments (The GNU C Library), c - How to write a "argv" and "argc" - Stack Overflow
