Why is putchar(tolower(ch)) printing out two of each character?

Question

This is my code, and each letter of the output is being printed twice. It is not the text file and it only happens when I insert my putchar(tolower) statement in but it is formatted exactly as it should be. What is wrong with the statement?

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

#define MAX_STRING_SIZE 20
#define MAX_LIST_SIZE 50

int readFile(char *filename); /* function declaration for readFile, defined below */
void punct();

/* main function */
int main(int argc, char *argv[]) {
    if (argc < 2) {
        printf("%s: usage %s textFileName \n", argv[0], argv[0]);
        exit(1);
    }

    readFile(argv[1]);

    return 0;
}

int readFile(char *filename) {
    char ch;
    FILE *fPtr;

    fPtr = fopen(filename, "r"); /*open file filename, r is read only */
    if (!fPtr) {
        return 0;
    }

    while ((ch = fgetc(fPtr)) != EOF) {
        putchar(tolower(ch));
        printf("%c", ch);
    }

    fclose(fPtr);

    return 1; /* because success */
}

You call `putchar` and then `printf`... – ThreeFx Jan 29 '17 at 23:24 — ThreeFx, Jan 29 '17 at 23:24
Because of this: `putchar(tolower(ch)); printf("%c", ch);` – tonypdmtr Jan 29 '17 at 23:24 — tonypdmtr, Jan 29 '17 at 23:24

score 5 · Answer 1 · answered Jan 29 '17 at 23:24

each letter of the output is being printed twice

This two lines:

putchar(tolower(ch));
printf("%c", ch);

Print two times the same character.

putchar puts a character to the standard output.
printf puts a formatted text to the standard output. In your case, the formatted text is the same character printed the line immediately above.

it only happens when I insert my putchar(tolower) statement in

That's because doing that you are actually doubling the output.

chqrlie · Answer 2 · 2017-02-04T19:37:51.417

There are multiple problems in your code:

ch must be defined with type int to accommodate all possible return values from fgetc(). These are all values of type unsigned char plus the special value EOF. Storing into a char variable poses multiple problems:
- The value EOF can no longer be tested accurately. If the type char is unsigned, (ch = fgetc(fp)) == EOF is always false. Conversely, if type char is signed, (ch = fgetc(fp)) == EOF might be true for ch = '\377, a valid byte value.
- if char is signed, tolower(ch) has undefined behavior for negative values.
each byte read from the file is output twice: first by putchar(tolower(ch)); and a second time by printf("%c", ch);.

Here is a corrected version:

#include <ctype.h>
#include <errno.h>
#include <stdio.h>
#include <string.h>

char *progname;    

int readFile(const char *filename) {
    int ch;
    FILE *fPtr;

    fPtr = fopen(filename, "r"); /* open file filename for reading */
    if (!fPtr) {
        fprintf(stderr, "%s: cannot open %s: %s\n",
                progname, filename, strerror(errno));
        return 1;  /* failure */
    }

    while ((ch = fgetc(fPtr)) != EOF) {
        putchar(tolower(ch));
    }
    fclose(fPtr);

    return 0;  /* success */
}

int main(int argc, char *argv[]) {
    progname = argv[0];
    if (argc < 2) {
        fprintf(stderr, "usage: %s textFileName\n", progname);
        return 1;
    }
    return readFile(argv[1]);
}

Why is putchar(tolower(ch)) printing out two of each character?

2 Answers2