Fopen Fails to read from Command line - c

Question

Fopen return NULL based on
'fopen' return a NULL
now I have one file to read and I want to output on console but it throws an error as fp return NULL here a snippet
#include<stdio.h>
int main(char argc,char **argv)
{
if(argc < 2)
{
printf("Usage:\t");
printf("Hash File path");
}
printf("Hash File : %s\n", argv[1]);
FILE *fp = fopen("argv[1]","r");
if(fp == NULL)
{
fprintf(stderr,"Can't open %s\n", argv[1]);
}
else
{
char buff[100];
while(!feof(fp))
{
if(fgets(buff,100,fp) == NULL) break;
fputs(buff,stdout);
}
fclose(fp);
}
return 0;
}
whenever I pass file Path E:\design\test.txt from the command line it always shows error report "Can't open...." like this
so the confusion is
why Fopen Failed to read from Command line?

Answer 1

There are few issues in code. Firstly, this
FILE *fp = fopen("argv[1]","r");
is wrong. First parameter of fopen() should be just argv[1] instead of "argv[1]". For e.g
FILE *fp = fopen(argv[1],"r");
if(fp == NULL) {
fprintf(stderr,"Can't open %s\n", argv[1]);
return 0; /* this you forgot, if fopen fails,then no need to proceed further */
}
Also if argc<2 is true then just wrinting printf() is not enough, you have to ask user to give correct command line input & don't let user proceed further. It should be
if(argc < 2) {
printf("Usage:\t");
printf("Hash File path");
return 0; /* return from here or use exit(0) */
}
Secondly, Read here why feof() is wrong. You can use it like
while(fgets(buff,100,fp) != NULL) { /* no need to use break */
fputs(buff,stdout);
}
instead of
while(!feof(fp))
{
if(fgets(buff,100,fp) == NULL) break;
fputs(buff,stdout);
}