C program complies but screen on console remains black

C program complies but screen on console remains black



I was writing a C program that would read and merge 3 files together (program not complete yet), however, as I was testing I realized the program compiles but the screen on the console remains blank!



Any help is appreciated, especially why is it blank?


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

int main()

printf("test");
//open three files for merging
FILE *fp1 = fopen("american0.txt","r");
FILE *fp2 = fopen("american1.txt","r");
FILE *fp3 = fopen("american2.txt","r");

printf("test");
//open file to store the result
FILE *fpm = fopen("words.txt", "w");

//creating an array to save the files data
char c;
char mergedFile[50];

//checking to make sure files are being read
if(fp1 == NULL && fp2 == NULL && fp3 == NULL && fpm == NULL)

printf("Could not open one or all of the files.n");
printf("Exiting program!");
exit(0);

printf("test");
//initializing counter values

//inserting data from file into an array
while ((c = fgetc(fp1)) != EOF)

fputc(c, mergedFile);

while ((c = fgetc(fp2)) != EOF)

fputc(c, mergedFile);

while ((c = fgetc(fp3)) != EOF)

fputc(c, mergedFile);


printf("%s",mergedFile[0]);
printf("test");

return 0;






fputc requires a FILE* as the second argument. How on earth did you get this to even compile when passing mergedFile, declared as char mergedFile[50]; to those function calls ? Shouldn't those all use fpm ? From what I see, mergedFile is pointless to have in this code at all. Removing it and all it's usages would also fix printf("%s",mergedFile[0]);, which makes no sense itself, as %s requires a const char*, and you're passing a char to that printf invoke.

– WhozCraig
Sep 15 '18 at 16:05



fputc


FILE*


mergedFile


char mergedFile[50];


fpm


mergedFile


printf("%s",mergedFile[0]);


%s


const char*


char


printf






Well I compiled and ran it and it did print stuff out, not sure what might be going wrong on your end. Don't take this wrong, as this feels like a stupid question to ask, but did you actually run it after compiling? Compiling will not execute the program, it will just produce a program that you can execute.

– pstrjds
Sep 15 '18 at 16:10







I feel like it is something minor and stupid that I'm doing that's causing this, However, I compile the program, all the warnings print out and when I run it the using "./a.out" it doesn't print out anything

– mahdi
Sep 15 '18 at 16:32







I also just put the printf("%s", mergeFile[0]); as a test statement which I realized was wrong, but thanks for pointing out the issue with the mergedFile in the while statements :)

– mahdi
Sep 15 '18 at 16:34






I believe the issue was with me passing mergedfile in the while loops, thank you for your help

– mahdi
Sep 15 '18 at 16:38




1 Answer
1



Error --> fputc requires a file pointer as it's second argument rather than an array: int fputc ( int character, FILE * stream );


fputc


int fputc ( int character, FILE * stream );



Points to be taken care of:


char



Here is a minimal corrected version:


#include <stdio.h>
#include <stdlib.h>
#define MAX 1000 //ADDED NEW

int main()

//open three files for merging
FILE *fp1 = fopen("american0.txt","r");
FILE *fp2 = fopen("american1.txt","r");
FILE *fp3 = fopen("american2.txt","r");

//open file to store the result
FILE *fpm = fopen("words.txt", "w");



//creating an array to save the files data
int c;
int i=0;
char mergedFile[MAX]=0; //MODIFIED & INITIALIZED

//checking to make sure files are being read
if(fp1 == NULL && fp2 == NULL && fp3 == NULL && fpm == NULL)

printf("Could not open one or all of the files.n");
printf("Exiting program!");
exit(0);

//initializing counter values

//inserting data from file into an array
while (((c = fgetc(fp1)) != EOF)&&(i<MAX)) //MODIFIED

mergedFile[i++]=c; //MODIFIED

while (((c = fgetc(fp2)) != EOF)&&(i<MAX)) //MODIFIED

mergedFile[i++]=c; //MODIFIED

while (((c = fgetc(fp3)) != EOF)&&(i<MAX)) //MODIFIED

mergedFile[i++]=c; //MODIFIED


printf("%s",mergedFile); //MODIFIED

return 0;







Better to use int c; than char c; to properly handle the typical 256 + 1 different results from fgetc().

– chux
Sep 15 '18 at 17:14


int c;


char c;


fgetc()






@chux thanks, corrected that!

– Observer
Sep 15 '18 at 17:17






I know you were copying OP's code, but I think that if statement checking for file opening success should be || not && since there will be a problem later if any file didn't open, not just if all didn't open.

– pstrjds
Sep 15 '18 at 18:41


if


||


&&






@user10334659 - I know that is what the OP has written (which is why I prefaced my comment with that), but since you are fixing the code I figured it was worth pointing out. As written it will only be true if all of the file open calls fail. If even one of them is successful the code will continue, which will cause problems later.

– pstrjds
Sep 15 '18 at 18:49






Thank you so much for all your help. this cleared up a lot of confusion!

– mahdi
Sep 16 '18 at 3:42



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