CODEWITHSUNDEEP
cprintfeoffgetcgetc
Prz3m3k
Prz3m3k

Reputation: 605

fgetc is always returning value 1

there is the following function:

void readAndPrint(FILE * f) {
    int c;

    while(c = fgetc(f) != EOF) {
        printf("%d", c);
    }
}

In the main() body I used the following code to use the above function:

FILE * pFile;

pFile=fopen ("myfile.txt","r");

readAndPrint(pFile)

;

Whatever I put into myfile.txt, the program prints out ones. For example, for abc, 111 is printed out.

I know that c in the function should be declared int to properly compare it to EOF. Also, I expected an int code from the ASCII set for each char in the text file to be printed out (97 for a, ...). I cannot figure out why it prints out 'ones'... Does you know the reason why? Thank you in advance.

Upvotes: 3

Views: 3126

Answers (1)

rashok
rashok

Reputation: 13424

(c = fgetc(f) != EOF) - Here first fgetc(f) != EOF this condition is happening and the result 1 or 0 is assigned to c. Always a condition check returns TRUE(1) or FALSE (0).

Do while((c = fgetc(f)) != EOF)

Upvotes: 13

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