How to read numbers separated by space using scanf

Question

I want to read numbers(integer type) separated by spaces using scanf() function. I have read the following:

It doesn't help me much.

How can I read numbers with space as delimiter. For e.g. I have following numbers as input 2 5 7 4 3 8 18 now I want to store these in different variables. Please help.

[link](http://stackoverflow.com/a/3765075/1317740) . Will that help? — NT_, May 03 '12 at 06:09
That is for reading white spaces. I want to read numbers separated by white spaces. — Jainendra, May 03 '12 at 06:13
Is it for a dynamically length string or a fixed set of numberss? — NT_, May 03 '12 at 08:01
Then why bothering with all the jumble code. Follow that [Answer](http://stackoverflow.com/a/10426060/1317740) . That's what you want then. It's not working for you?. — NT_, May 03 '12 at 08:44

Nandkumar Tekale · Accepted Answer · 2012-05-03T06:32:20.527

40

I think by default values read by scanf with space/enter. Well you can provide space between '%d' if you are printing integers. Also same for other cases.

scanf("%d %d %d", &var1, &var2, &var3);

Similarly if you want to read comma separated values use :

scanf("%d,%d,%d", &var1, &var2, &var3);

edited May 03 '12 at 06:32

answered May 03 '12 at 06:13

Nandkumar Tekale

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2

scanf ignores whitespace with most `%` directives, so you generally don't want spaces in your format strings. So for the first case (whitespace delimited integers), you want `scanf("%d%d%d", &var, &var2, &var3);` The extra spaces don't actually hurt anything here (they have no effect), but in other cases you don't want them unless you need them. – Chris Dodd Feb 07 '13 at 21:23
Why do you have to use &var1 here (is var1 an int type variable?)? Isn't that just a 'label' for var1's memory adress? – YoTengoUnLCD Oct 04 '15 at 23:24
@YoTengoUnLCD yes exactly. The `scanf` function needs the address in order to overwrite the value at that address. This is true for `ints` and any other type you might use with `scanf`. There's nothing magic going on here, if you have any function that needs to overwrite a value, that function needs the address, not the value itself. – parker.sikand Jul 21 '16 at 16:48

score 7 · Answer 2 · answered Feb 07 '13 at 21:20

scanf uses any whitespace as a delimiter, so if you just say scanf("%d", &var) it will skip any whitespace and then read an integer (digits up to the next non-digit) and nothing more.

Note that whitespace is any whitespace -- spaces, tabs, newlines, or carriage returns. Any of those are whitespace and any one or more of them will serve to delimit successive integers.

score 3 · Answer 3 · answered May 31 '12 at 02:14

3

int main()
{
char string[200];
int g,a,i,G[20],A[20],met;

gets(string);
g=convert_input(G,string);

for(i=0;i<=g;i++)
    printf("\n%d>>%d",i,G[i]);
return 0;
}

int convert_input(int K[],char string[200])
{
int j=0,i=0,temp=0;
while(string[i]!='\0')
{
    temp=0;
    while(string[i]!=' ' && string[i]!='\0')
        temp=temp*10 + (string[i++]-'0') ;
    if(string[i]==' ')
        i++;
    K[j++]=temp;
}
return j-1;
}

answered May 31 '12 at 02:14

niraj.nijju

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Would come kind of description not help? – Bridge May 31 '12 at 08:06

score 2 · Answer 4 · edited Mar 30 '15 at 18:36

2

It should be as simple as using a list of receiving variables:

scanf("%i %i %i", &var1, &var2, &var3);

edited Mar 30 '15 at 18:36

Alexander

59,041
12
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answered May 03 '12 at 06:07

what is fp here? In the above case the delimiter is \n, I want to use space as delimiter. – Jainendra May 03 '12 at 06:10
Sorry, I removed it since it is not pertinent in this example. You don't need to specify the space delimiter somewhere, it is implicit in your format string. – May 03 '12 at 06:36
Never use `%i`.. unless you know what it does. – Antti Haapala -- Слава Україні Sep 15 '19 at 18:57

score 0 · Answer 5 · answered Dec 16 '21 at 11:05

With this solution, it's possible to read positive and negatives integers:

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

#define BUFFER_SIZE 50


int convert_input (int * v, char * buffer) {
    int len = 0, i = 0, temp = 0, positive_or_negative_one = 1;
    while(buffer[i]!='\0') {
        temp = 0;

        if (buffer[i] == '-'){
            positive_or_negative_one = -1;
            i++;
        } else {
            while(buffer[i] != ' ' && buffer[i] != '\0')
                temp = temp*10 + (buffer[i++]-'0');

            if(buffer[i]==' ')
                i++;
            v[len++] = temp * positive_or_negative_one;
            positive_or_negative_one = 1;
        }
    }
    return len;
}

int main(int argc, char const *argv[]) {
    int *a = NULL;
    int count_a, len=0;
    char buffer[BUFFER_SIZE];

    printf("Input numbers here: ");
    gets(buffer);
    for (int i = 0; i < strlen(buffer); i++) {
        if (buffer[i] == ' '){
            len+=1;
        }
    }

    a = (int*) malloc(sizeof(int) * len + 1);
    count_a = convert_input(a, buffer);

    for (int i = 0; i < count_a; i++) {
        printf("%d\n", a[i]);
    }

    free(a);

    return 0;
}

Input and output example:

Input numbers here: 1 2 3 -4 10
1
2
3
-4
10

How to read numbers separated by space using scanf

5 Answers5

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