Question

1 #include <stdio.h>

2 #include <stdlib.h>

3

4 extern char **environ;   

5 void output(char *a[], char *b[]) {

6 int c = atoi(a[0]);

7 for (int i = 0; i < c && b[i]; ++i) {

8 printf("%s", b[i]+2);

9 }

10 }

11

12 void main(int argc, char *argv[]) {

13     

14 switch (argc) {

15 case 1:

16 for (int i = 0; environ[i]; ++i) {   

17 printf("%s\n", environ[i]);

18 }

19 break;

20 default:

21 output(argv + 1, environ + 2);

22 exit(argc);

23 }

24 }

After executing the command $ ./m0 (assuming that the code above was compiled with $ gcc m0.c -o m0), suppose that the output is

A=9

B=4

C=1

D=4

E=5

What is the output of executing this command $ ./m0 2 3 4 5?

(atoi(str) converts the string argument str to an integer)

Answer 1

The program prints 14 and exits with exit code 5.

Explanation:
argv contains items passed on the command line, i.e. the array of strings m0,2,3,4,5.
argv + 1 points to the array 2,3,4,5 which is what gets passed to output as the first parameter.
Looking at the first run, the external array environ contains the strings "A=9","B=4","C=1", "D=4","E=5", null
environ + 2 therefore points to "C=1", "D=4","E=5", null
In output, the parameters are (a = {"2", "3", "4", "5"}, b = {"C=1", "D=4","E=5", null})
Therefore c in line 7 will contain 2 and the loop will run two times
The first time, it will look at "C=1" and print starting from offset 2, i.e "1"
The second time, it will look at "D=4" and print starting from offset 2, i.e. "4"
Since there is no space or any other characters printed, the output shows up as 14