Hi.. Could anyone tell me what I did wrong? I kept getting segmentation faults.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
FILE *fp;
int num;
char *name;
fp = fopen (argv[1],"r");
while (!feof(fp)) {
fscanf (fp, "%d\t%s", &num, name);
printf ("num=%d name=%s\n", num, name);
}
fclose (fp);
return 0;
}
ie. fp contains data such as:
123 abc
456 def, etc..
And I want to put them into arrays.
>>while (!feof(fp))
this is wrong, read this
i thought you had to type **argv[] in the main function's parameters yours is missing the []. I could be wrong though
You've got one to many *.Originally posted by killjoyntk
i thought you had to type **argv[] in the main function's parameters yours is missing the []. I could be wrong though
int main( int argc, char *argv[] )
Quzah.
Hope is the first step on the road to disappointment.
i thought about that after i posted that, I always used *argv[] .
Ive used char **argv many times, why is it wrong??Originally posted by quzah
You've got one to many *.
int main( int argc, char *argv[] )
Quzah.
What is the difference between char *argv[] and char **argv?
if you access an array like its a pointer then it does not really matter right?
Code:int main(int argc, char **argv) { int i; if(argv < 2) { prinf("need at least more then one argument"); return(0); } for(i=argc; i > 0; i--) { printf("%s", **(argv+i)); } return(0); }
Last edited by squid; 04-18-2003 at 11:43 PM.
I wasn't saying that was wrong.Originally posted by squid Ive used char **argv many times, why is it wrong??[/code]
Read my post over again. See what I quoted? "char **argv[]" is wrong. Like I said, one too many * in that.
Quzah.
Hope is the first step on the road to disappointment.
Oops I missed that. In that case you are right. OOPS
