C program doesn't print text when trying to access string data using pointers

Question

I'm going through "C Programming: A Modern Approach" by K.N. King right now. This program is taken from "Section 12 - Pointers and Arrays", Project 1b. Roughly, the goal is: Write a program that reads a message, then prints the reverse of the message. Store the message in an array, and keep track of the current position in the array using a pointer.

Currently, I'm just trying to get the message to print normally before I try and make it print backwards. I'm frustrated because my printMessage() function is 1:1 with the print_message() function from here, however when the program is run and some text is entered, it just prints out a blank line. This happens if the line within the for-loop of printMessage() is changed to 'printf("%c", *p);' as well.

Here is my code:

#include <stdio.h>

#define MAX_MESSAGE_LENGTH 100

int readMessage(char msg[]);
void printMessage(char msg[], int len);

/* finish later
 void printMessageReversed(char msg[], int len); */

int main(void){
    int msg_length;
    char c[MAX_MESSAGE_LENGTH];

    printf("Enter a message: ");
    msg_length = readMessage(c);

    printf("Length: %d", msg_length);

    printf("\nMessage: ");
    printMessage(c, msg_length);

    return 0;
}

int readMessage(char msg[]){
    int i;
    for(i = 0; (*msg = getchar()) != '\n'; i++);
    return i;
}

void printMessage(char msg[], int len){
    char *p;

    for(p = msg; p < msg + len; p++){
        putchar(*p);
    }
}

Wondering if anyone can help identify my problem? I'm sure it's something painfully obvious or naive that I'm just missing. Thanks for any help.

@Shravan40: No, that's not it. The msg parameter is a pointer. — Keith Thompson
– Keith Thompson, Commented Sep 9, 2016 at 19:44
@Shravan40: C does not support pass by reference! It is strictly pass by value. — too honest for this site
– too honest for this site, Commented Sep 9, 2016 at 19:45
@Olaf: Of course -- but passing a pointer (by value) can give you the equivalent functionality of pass-by-reference. — Keith Thompson
– Keith Thompson, Commented Sep 9, 2016 at 19:46
Apart from the lack of increment others have commented, you have not written a nul terminator at the end of the input to make the array a "string", in readMessage — Weather Vane
– Weather Vane, Commented Sep 9, 2016 at 19:47
@Olaf : In C, Pass-by-reference is simulated by passing the address of a variable (a pointer) and dereferencing that address within the function to read or write the actual variable. This will be referred to as "C style pass-by-reference." — Shravan40
– Shravan40, Commented Sep 9, 2016 at 19:47

Keith Thompson · Accepted Answer · 2016-09-09 20:35:22Z

9

int readMessage(char msg[]){
    int i;
    for(i = 0; (*msg = getchar()) != '\n'; i++);
    return i;
}

Each character is assigned to *msg (equivalently to msg[0]). The msg pointer is never updated and never indexed by i, so you never assign values to the remaining characters in the array whose initial element is pointed to by msg.

Replacing *msg by msg[i] is one partial solution. Note that this still doesn't write a null '\0' terminator into msg, so the msg array does not contain a string. But since your printMessage function doesn't require a string (it takes a len argument), that's not strictly necessary. Still, if you're going to operator on the array in other ways, it's probably a good idea to null-terminate it.

Also, as chux points out in a comment, you have an infinite loop if getchar() returns EOF before returning '\n' (on Unix that can happen if you type Ctrl-D twice in the middle of a line, or if you're reading from a file), and it has no protection against a buffer overrun.

edited Sep 9, 2016 at 20:35

answered Sep 9, 2016 at 19:44

Keith Thompson

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8 Comments

pm100 Over a year ago

so you need msg[i] = getchar()

Sato Over a year ago

Thank you! Altered the for-loop to read for(c = msg; (*msg = getchar()) != '\n'; msg++, i++); and it works perfectly now.

Sato Over a year ago

@pm100 and yes that would also work just as well, thanks :)

Sato Over a year ago

Also realise the example I just gave is probably overcomplicated so I'll try and cut it down a bit.

chux Over a year ago

Side issues: for(i = 0; (*msg = getchar()) != '\n'; i++); is potential infinite loop upon reading EOF and it has no buffer over-run protection.

|

J3K · Accepted Answer · 2016-09-09 20:03:06Z

1

Use it like that :

int readMessage(char msg[]){
    int i;
        for (i = 0; (*(msg + i) = getchar()) != '\n'; i++);
    return i;
}

answered Sep 9, 2016 at 20:03

J3K

194 bronze badges

3 Comments

Sato Over a year ago

This is far more concise than everything I was trying. Thanks!

Sato Over a year ago

@WeatherVane agreed, but for the purposes of this chapter of the book I'm going to try and stick entirely to pointers, just so I can get used to them.

Weather Vane Over a year ago

@Sato yeah, but the argument was an array (although that decays to a pointer). Another point: please be aware that if the argument was int msg[] instead of char msg[] there is no size adjustment required when using *(msg + i) = getchar() - just as with array indexing, the pointer arithmetic takes care of the size for you.

Niklas Rosencrantz · Accepted Answer · 2016-09-09 19:56:41Z

0

Write a program that reads a message, then prints the reverse of the message

#include <stdio.h>
#define MAX_MESSAGE_LENGTH 100
int main(void) {
    int i, j;
    char c[MAX_MESSAGE_LENGTH];
    printf("Enter a message: ");
    fgets(c, sizeof c, stdin);
    printf("Reverse Order\n");
    i = 0;
    while (c[i] != '\0') {
        if (c[i] == ' ' || c[i] == '\n') {
            for (j = i - 1; j >= 0 && c[j] != ' '; j--)
                printf("%c", c[j]);
            printf(" ");
        }
        i++;
    }
    printf("\n");
    return 0;
}

Test

Enter a message: Stackoverflow

Reverse Order
wolfrevokcatS

Try it with ideone.

edited Sep 9, 2016 at 19:56

answered Sep 9, 2016 at 19:50

Niklas Rosencrantz

26.7k78 gold badges247 silver badges447 bronze badges

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