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Homework and Emergencies Homework & Coursework Questions trouble understanding file option and command line arguments Post 302558629 by heywoodfloyd on Saturday 24th of September 2011 03:01:08 PM
Old 09-24-2011
trouble understanding file option and command line arguments

Hi,

I am creating a program with the C language that simulates the WC command in Unix. My program needs to count lines, bytes and words. I have not added the code to count bytes and words yet. I am having trouble understanding what the file option/flag '-' does. I can not visualize how it moves through program code. argc and argv look at one line at a time and they only look at what is on the command line. I am not sure how a program is broken up into files for WC to read.

this is the code I have so far:

Code:
/* wc simulate */

#include <stdio.h>
#include <stdlib.h>

char *pgmname; /* name of this program */

int line_count = 0;
int word_count = 0;
int byte_count = 0;

FILE *fp;

void main(argc, argv)

int argc; char *argv[];
{


int i; 
char *cp;

pgmname = argv[0];
fp = stdin;

for(i = 1; i < argc; i++) {
        cp = argv[i];
        if(*cp == '-'){
                if(*++cp == '\n'){
                        line_count++;}
                
                
        }

        else {
                  if(fp != stdin) {
                   fprintf(stderr, "%s: too many arguments\n", pgmname);
                   exit(1);
                }

                fp = fopen(cp, "r")
                if(fp == NULL) {             
                   fprintf(stderr, "%s: unable to read %s\n", pgmname, cp);
                   exit(1);
                }

        }

                printf("%d\n", line_count);
}

                

}

Right now I have code to just count the lines. I am testing the program on a hello world program below:

Code:
#include <stdio.h>

main()
{
printf("Hello World!\n");
}

when I test the program, I get 0 for line count.

California State University, Northridge, USA, Prof Gabrovsky, Comp 322
 

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BSWAP(3)						     Linux Programmer's Manual							  BSWAP(3)

NAME
bswap_16, bswap_32, bswap_64 - reverse order of bytes SYNOPSIS
#include <byteswap.h> bswap_16(x); bswap_32(x); bswap_64(x); DESCRIPTION
These macros return a value in which the order of the bytes in their 2-, 4-, or 8-byte arguments is reversed. RETURN VALUE
These macros return the value of their argument with the bytes reversed. ERRORS
These macros always succeed. CONFORMING TO
These macros are GNU extensions. EXAMPLE
The program below swaps the bytes of the 8-byte integer supplied as its command-line argument. The following shell session demonstrates the use of the program: $ ./a.out 0x0123456789abcdef 0x123456789abcdef ==> 0xefcdab8967452301 Program source #include <stdio.h> #include <stdint.h> #include <stdlib.h> #include <inttypes.h> #include <byteswap.h> int main(int argc, char *argv[]) { uint64_t x; if (argc != 2) { fprintf(stderr, "Usage: %s <num> ", argv[0]); exit(EXIT_FAILURE); } x = strtoul(argv[1], NULL, 0); printf("0x%" PRIx64 " ==> 0x%" PRIx64 " ", x, bswap_64(x)); exit(EXIT_SUCCESS); } SEE ALSO
byteorder(3), endian(3) COLOPHON
This page is part of release 4.15 of the Linux man-pages project. A description of the project, information about reporting bugs, and the latest version of this page, can be found at https://www.kernel.org/doc/man-pages/. Linux 2017-09-15 BSWAP(3)
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