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void main(int argc,char *argv[],char *env[])
{
int i;
for(i=1;i<argc;i++)
printf("%s",env[i]);
}
- 3 Answers
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Answer / gaurav
It’s interesting.
This program reads environmental variables. It is as same as
"env" command in unix.
But program is not proper. i.e. if you pass n arguments to
program, then it reads n environmental variables only.
Try this program
void main(int argc,char *argv[],char *env[])
{
int i;
if (2 <=argc){
for(i=0;i<atoi(argv[1]);i++)
printf("\n%s",env[i]);
}else printf("\nPlease enter no. of env variables you want
e.g. 'a.out 5'\n");
}
| Is This Answer Correct ? | 4 Yes | 1 No |
Answer / siddique
void main(int argc, char *argv[], char *env[])
{
int i=0;
while(env[i])
{
printf("%s\n", env[i]);
i++;
}
}
| Is This Answer Correct ? | 2 Yes | 0 No |
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C 
