The number of inputs I have to enter is one more than required for the output to display

Question

I have to enter one more input than required for the code to start running. For example, for 2 strings I have to enter 3 strings, only then the output shows. Please help.

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

int main () {
    int num_strings;
    char arr[num_strings][100];
    scanf("%d",&num_strings); 
    for (int i=0;i<=num_strings-1;i++) {
        scanf("%s\n",&arr[i]);
        
    }
    for (int j=0;j<=num_strings-1;j++) {
        if (strlen(arr[j])>10) {
            printf("%c%d%c",arr[j][0],strlen(arr[j])-2,arr[j][strlen(arr[j])-1]);
            printf("\n");
        } else {
            printf("%s",arr[j]);
            printf("\n");
            
        }
        
    }
    
}

`scanf("%s\n",&arr[i]);` should be `scanf("%s",&arr[i]);`, `scanf()` keeps on reading and ignores after a non whitespace is found. — alex01011, Mar 31 '22 at 18:56
Aside: `i<=num_strings-1` is more classically understood as `i < num_strings`. — Oka, Mar 31 '22 at 18:58
Thanks Alex, yeah it does very nicely, also Oka input noted just beginning out. — Arnav Goyal, Mar 31 '22 at 19:03
When you declare the 2D array 'char arr[num_strings][100];', what value will the generated code use for 'num_strings'? — Martin James, Mar 31 '22 at 19:07

Oka · Answer 1 · 2022-03-31T19:36:15.503

The following

int num_strings;
char arr[num_strings][100];
scanf("%d",&num_strings);

is out of order. Changing the value of num_strings does not retroactively adjust the size of arrays that were initialized using this variable. As such, arr is initialized with an indeterminate value, since num_strings is uninitialized. This is an example of invoking Undefined Behavior.

&arr[i] is of type char (*)[100], you simply want arr[i], which will decay to a char * (the correct type for %s) when passed to scanf.

Remove the trailing whitespace from the scanf format, or it will hang until it reads non-whitespace ("%s\n" -> "%s").

Limit the length of the strings you read with a field-width specifier as to not overflow the buffers (%99s).

strlen returns a size_t, the format specifier for printing this is %zu.

Note that scanf can fail for a variety of reasons. Consider checking that the return value of each call was (or was not) the expected number of conversions, e.g.,

if (2 != scanf("%d%d", &a, &b))
    /* failure */;

and handle any failures.

Here is a functional version of your program, without error checking:

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

int main(void) {
    int num_strings;

    scanf("%d", &num_strings);

    char arr[num_strings][100];

    for (int i = 0; i < num_strings; i++)
        scanf("%99s", arr[i]);

    for (int j = 0; j < num_strings; j++) {
        if (strlen(arr[j]) > 10) {
            printf("%c%zu%c\n",
                    arr[j][0], strlen(arr[j]) - 2,
                    arr[j][strlen(arr[j]) - 1]);
        } else {
            printf("%s\n", arr[j]);
        }
    }

}

stdin:

3
hello
thisisalongerstring
world

stdout:

hello
t17g
world

The number of inputs I have to enter is one more than required for the output to display

1 Answers1