Guest_imported
New member
- Jan 1, 1970
- 0
What does this do...
int main(int argv, char **argc) {
char buf[256];
strcpy(buf,argc[1]);
exit(1);
}
int main(int argv, char **argc) {
char buf[256];
strcpy(buf,argc[1]);
exit(1);
}
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Just curious
- Thread starter Guest_imported
- Start date
- Status
Eddie:
All it does is copy your first command line argument into character variable buffer.
By convention (although you've done nothing wrong syntax wise), argc is the argument count and argv is the pointer to a char pointer:
void main(int argc, char **argv) {
char buf[256];
if(argc != 2)
{
printf("arg count error\n"
;
exit(1);
}
strcpy(buf,argv[1]);
printf("buf is %s\n", buf);
exit(0);
}
Regards,
Ed
All it does is copy your first command line argument into character variable buffer.
By convention (although you've done nothing wrong syntax wise), argc is the argument count and argv is the pointer to a char pointer:
void main(int argc, char **argv) {
char buf[256];
if(argc != 2)
{
printf("arg count error\n"
;exit(1);
}
strcpy(buf,argv[1]);
printf("buf is %s\n", buf);
exit(0);
}
Regards,
Ed
- Status