How to check if String is empty in C

Question

I have a void function which take the string from a global variable and prints them.

The problem is that the void function prints a message/does stuff even if the string is empy. so if the global variable is empt it just prints "message is:" i only want to do something if the string has characters.

what i want to do is to check if the string is not empty, then print a message.

// global variable

char msg[30];
scanf("%s", &msg);

void timer(void) {
    //i want to only print if there is a message
    printf("message: %s", msg);
}

thanks for any help

@juanchopanza are we sure that strln wouldn't return 30 here ? — 4rlekin
– 4rlekin, Commented Oct 28, 2014 at 13:22
@4rlekin Then the string wouldn't be empty. I am assuming a valud null-terminated string. Otherwise the problem is a different one. — juanchopanza
– juanchopanza, Commented Oct 28, 2014 at 13:23
@juanchopanza i meant that provided that scanf() didn't worked uninitialized array would contain 30 * sizeof(char) bytes of garbage AFAIK Also return from scanf() idicates failure/success — 4rlekin
– 4rlekin, Commented Oct 28, 2014 at 13:25

Edward · Accepted Answer · 2014-10-28 13:43:55Z

16

The easy way to do it would be like this:

if (msg[0])
    printf("message: %s", msg);

If, at some later date, msg is a pointer, you would first want to assure it's not a NULL pointer.

if (msg && msg[0])
    printf("message: %s", msg);

A test program to demonstrate:

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

char *msg;

void test() {
    if (msg && msg[0])
        printf("string is %s\n", msg);
}

int main()
{
    msg = NULL;
    test();
    msg = calloc(10, 1);
    test();
    msg[0] = 'm';
    msg[1] = 'e';
    test();
    free(msg);
}

On my machine the output is:

string is me

edited Oct 28, 2014 at 13:43

answered Oct 28, 2014 at 13:23

Edward

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5 Comments

Amit Sharma Over a year ago

I think there is no difference between msg && msg[0] and msg[0] ??

Edward Over a year ago

If it's defined as char *msg the difference is that msg[0] refers to the first character and msg refers to the pointer itself.

Amit Sharma Over a year ago

You can always write msg as msg[0] because it will treated as *(msg+0) which is exactly the same as msg. Array name is just a constant pointer .So how it make difference, here

Baldrickk Over a year ago

@AmitSharma no. msg evaluates to the address of the first char. msg[0] is the same as *(msg+0) but not msg. msg && msg[0] is true when the pointer is: not null. references a char with a non-zero value. It also short-circits, so it won't try to dereference a null pointer.

Edward Over a year ago

@AmitSharma: I've added a silly little test program to the answer. Try it.

unwind · Accepted Answer · 2014-10-28 13:39:15Z

11

Strings in C are represented as character arrays, terminated by a character whose value is zero (often written as '\0'). The length of a string is the number of characters that come before the zero.

So, a string is empty if the first character in the array is zero, since then by definition no characters came before the zero.

This can be checked like so:

if(msg[0] != '\0')
{
  /* string isn't empty! */
}

This is very explicit code, the shorter if(msg[0]) or if(*msg) has the exact same semantics but can be slightly harder to read. It's mostly a matter of personal style.

Note though that in your case, when you use scanf() to read into msg, you should check the return value from scanf() before checking the contents of msg. It's perfectly possible for scanf() to fail, and then you can't rely on msg being set to an empty string.

edited Oct 28, 2014 at 13:39

answered Oct 28, 2014 at 13:33

unwind

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8 Comments

ryyker Over a year ago

+1 for mentioning return of scanf(). (nobody else saw that as an issue.)

unwind Over a year ago

@ryyker Nobody ever does, around here.

Edward Over a year ago

One could also note that the use of a global variable is probably not a good idea either. Seemed orthogonal to the question asked, though, so I didn't include it in my answer.

ryyker Over a year ago

@Edward - true especially for embedded, but outside of embedded, there are places where globals are useful. eg. extern scoped struct arrays provide simplicity, and the cost is very affordable when memory is so cheap. (for applications running on a desktop, for example)

Edward Over a year ago

@ryyker: While there may well be instances where a global variable is needed, this is certainly not one of them.

|

Rizier123 · Accepted Answer · 2014-10-28 13:53:44Z

3

This should work for you:

The Message will only be printed if it's longer then 0.

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

void timer() {

    char msg[30];
    scanf(" %s", msg);

    if(strlen(msg) != 0)
        printf("message: %s", msg);
}

int main() {

    timer();

    return 0; 
}

edited Oct 28, 2014 at 13:53

answered Oct 28, 2014 at 13:22

Rizier123

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3 Comments

Amit Sharma Over a year ago

This will give error. You have to include#include<string.h>.

Rudy Velthuis Over a year ago

I wonder why no one recommends checking the return value of scanf? If it fails, the msg array may still contain garbage, and the string length is not necessarily 0.

Spikatrix Over a year ago

Dosen't the & need to be removed from scanf?

MD XF · Accepted Answer · 2017-03-04 17:44:01Z

1

Here are all the ways I found (and can think of) to check if a string is empty. Sorted by efficiency.

msg[0] == '\0'
strlen(msg) == 0
!strcmp(msg, "")

And this, which should only happen if the user somehow avoided input (e.g. with Ctrl+Z) or there was an error

scanf("%s", &msg) <= 0

answered Mar 4, 2017 at 17:44

MD XF

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Comments

Gerald H · Accepted Answer · 2020-11-03 02:30:38Z

0

I think a quick and easy solution is to use strlen() in string.h to check the length of the string. if 0 then it is empty.

For example

// Check string is NOT empty.
if (strlen(current_word) > 0)
            {
                printf("%s %lu", current_word, strlen(current_word));
                strcpy(current_word, "");
                print_line();
            }

answered Nov 3, 2020 at 2:30

Gerald H

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Comments

Pranu Pranav · Accepted Answer · 2020-12-08 03:29:26Z

Using Strlen() function.
Comparing with NULL string using strcmp();
Checking first character is equal to '\0'.
If string is inputted using scanf then check for return value of scanf() function

1st method:

char msg[30];

scanf("%s", &msg);

void timer(void) {

if(strlen(msg)>0)

printf("message: %s", msg);

}

2nd method:

char msg[30];

scanf("%s", &msg);

void timer(void) {

if(strcmp(msg,"")!=0)

printf("message: %s", msg);

}

3rd method:

char msg[30];

scanf("%s", &msg);

void timer(void) {

if(msg[0]!='\0')

printf("message: %s", msg);

}

4th method:

char msg[30];

int rv=scanf("%s", &msg);

void timer(void) {

if(rv>0)

printf("message: %s", msg);

}

For your reference about c string

https://www.programiz.com/c-programming/c-strings

https://www.geeksforgeeks.org/strings-in-c-2/

https://scholarsoul.com/string-in-c/

https://www.guru99.com/c-strings.html

El Barrio 13 · Accepted Answer · 2014-11-04 02:05:51Z

-1

Ok,you can try this:

char msg[30];
scanf("%s",&msg);

void timer(void)
{
    //i want to only print if there is a message
    if(!msg.length == 0 )
        printf("message: %s", msg);
}

`

answered Nov 4, 2014 at 2:05

El Barrio 13

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