CODEWITHSUNDEEP
carraysarguments
Stamoulohta
Stamoulohta

Reputation: 659

array into arguments in c

Am I forced to use loops and realy long lines or is something like this possible in c?

int a[5] = {1, 2, 3, 5, 8};
printf("%d %d %d %d %d\n", a[0], a[1], a[2], a[3], a[4]); // this is ok but
printf("%d %d %d %d %d\n", a);                // is this somehow possible??

I am asking if there is a way.

Upvotes: 1

Views: 60

Answers (3)

user3920237
user3920237

Reputation:

This line:

printf("%d %d %d %d %d\n", a);

is actually just plain undefined behavior because 1) the format string expects 5 arguments, but you only provide one and 2) it expects a %d but you provide an array.

There isn't a built-in way to output the contents of an array, except for character arrays. For %s, it will print up to the terminating null byte or up to the maximum amount of characters specified by the precision. Even if you were able to provide the size of the array, the printf specification doesn't support it. Just use a loop.

Upvotes: 0

Vlad from Moscow
Vlad from Moscow

Reputation: 311018

Thy the following trick

#include <stdio.h>

int main(void) 
{
    struct { int a[5]; } A = { { 1, 2, 3, 5, 8 } };

    printf( "%d %d %d %d %d\n", A.a[0], A.a[1], A.a[2], A.a[3], A.a[4] );
    const char *f = "%d %d %d %d %d\n";
    printf( f, A );                 

    return 0;
}

The output is

1 2 3 5 8
1 2 3 5 8

Upvotes: -1

Sourav Ghosh
Sourav Ghosh

Reputation: 134336

Point 1.

printf("%d %d %d %d %d\n", a[0], a[1], a[2], a[3], a[4]);

Perfectly ok, but think of the horror when you have, say 10,000 elemnts.

Point 2.

printf("%d %d %d %d %d\n", a);

Problem with basic concept. You did not supply enough values to print.

Baseline: Stick to loops while printing array elements one by one.

Upvotes: 3

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