How to use the malloc() function for char string/array?

Question

I'm writing a code that reads in a list of characters (using redirection), but the string I'm storing the characters to has a size of 41. The number of characters do not come close to the size of the array, therefore I need to use malloc() in order to get the exact array size. When I print the array without using the malloc function is prints symbols and other gibberish along with the string. I need to malloc the string so the symbols and gibberish won't be included.

Code:

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

int main() {

    char c;
    char r1[41];
    int a = 0;

    while(scanf("%c", &c) != EOF) {
        if(c >= 'A' && c <= 'Z' || c == '[' || c == '.' || c == '!') {
        r1[a] = c;
        a++;
        }
    }

    printf("\nThe string is: %s\n", r1);

return 0;
}

Output:

The string is TR[!.DFQ▒ ▒ ▒ 這

Your string is an array of 42 pointers to char. You want char string[40+1];. (And of course you'll need to store some value in it.) — Keith Thompson
– Keith Thompson, Commented Nov 20, 2013 at 21:23
There is nothing in that code that reads any value(s) in from anywhere. Thus the contents of your string array are uninitialized, which leads to further undefined behavior... — twalberg
– twalberg, Commented Nov 20, 2013 at 21:28
Any reason why you're not using C++? With the string class you won't need to malloc anything. — SleepingSpider
– SleepingSpider, Commented Nov 20, 2013 at 21:31
I'm in a computer science class in college that is strictly C programming. — Tom Murphy
– Tom Murphy, Commented Nov 20, 2013 at 21:52

John Kugelman · Accepted Answer · 2013-11-20 21:59:32Z

2

There are a couple of points of confusion mixed together here.

The reason your code is printing gibberish isn't due to fixed-length vs. variable-length. Switching to malloc() won't fix that problem. Make sure you have a null terminator ('\0') at the end of the string:
```
while(scanf("%c", &c) != EOF) {
    ...
}

r1[a] = '\0';
```
Your original code mixed array and pointer syntax. For a single string you should use either char[] or char* but not both.

For a fixed-size string, declare it as:
```
char string[40+1];
```
For a variable-length string, do:
```
char *string = malloc(length + 1);

if (string == NULL) {
    fprintf(stderr, "malloc failed\n");
    exit(EXIT_FAILURE);
}

// memory allocation successful
```

edited Nov 20, 2013 at 21:59

answered Nov 20, 2013 at 21:25

John Kugelman

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4 Comments

Tom Murphy Over a year ago

I was using the declaration: char string[40+1] at first. I think it will be best if I show my entire code.

Tom Murphy Over a year ago

I debugged the program plenty of times, when I used the exact size of the array, the symbols and gibberish didn't print. So it is the extra indexes that causes the symbols and gibberish.

user529758 Over a year ago

@TomMurphy No, it isn't. It's the undefined behavior caused by the missing NUL-terminator. Be humble, don't try to be smarter than the person who pointed out and solved your obvious beginner mistake.

Tom Murphy Over a year ago

Thanks this works. My professors told me that I must allocate the memory for the string in order to get it to print correctly.

Keith Thompson · Accepted Answer · 2013-11-20 21:26:52Z

char *string[40+1];

This defines string as an array of 41 pointers to char. If you want it to hold a string, you'll want to define it as:

char string[40+1];

If you know (at run time, but not at compile time) how long the string will be, you can use malloc -- but you'll want to assign the result to a single pointer of type char*:

char *s = malloc(len + 1);
if (s == NULL) {
    /* malloc failed, handle the error */
}

This causes s to point to the first (0th) element of an array of char. That array is uninitialized, so you'll need to copy some data into it.

Once you've done that, and assuming the contents are properly null-terminated, then you can print the string's value using:

printf("The string is: \"%s\"\n", string);

The comp.lang.c FAQ is an excellent resource for this kind of thing.

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