What is the C++ equivalent of C language's scanf("%*c")?

Question

Objective :

• Simplest and similar way to discard any unwanted (which in my case is every non numeric characters but I'd like a solution for a more general case) and remove them from buffer. Example,

#include<stdio.h>
void main(void){
    int num1, num2;
    
    printf("Enter num1 here : ");
    scanf("%d%*[^\n]%*c", &num);//First scanf statement.
    
    printf("Enter num2 here : ");
    scanf("%d", &num2);//Second scanf statement.
    
    printf("num1  is : %d\n
            num2 is : %d", num, num2);
}

    /* OUTPUT */
    1st number : 25 asdfasfasfasf
    2nd number : 30
    Input 1 : 25
    Input 2 : 30
    /* This will work by discarding any non */
    /* In the program above I didn't write a printf statment to display the text "1st num, 2nd num". This is just so the output is a bit more clear. I am aware of it.  */

Now if you change the first scanf from scanf("%d%[^\n]%*c"); to scanf("%d"); and give the same input, you'd get the following output :

#include<stdio.h>

void main(void){
    int num1, num2;

    printf("Enter num1 here : ");
    scanf("%d", &num1);
    
    printf("Enter num2 here : ");
    scanf("%d", &num2);

    printf("num1 : %d\nnum2 : %d", num1, num2);
}

    //OUTPUT//
    Enter num1 here : 11 asdf
    Enter num2 here : num1 : 11
    num2 : 32764
   /*As you might see, I am not prompted for 2nd input. Instead the already present characters in the buffer are used.*/

Briefing :

• In C scanf("%d%[^\n]%*c"); would remove all the redundant characters like spaces, newline, alphanumeric that come after the number form the buffer before/ after taking another input. How can I achieve the same in C++ where after my cin >> var; would take the remaining characters in the buffer and then discard them. This only works for numbers from 0-9. But what I am interested in is the *[^\n] and *c because it helps me read the characters from buffer without saving them anywhere which technically means that they get discarded.

Exclusions :

• cin >> ws;
• cin.ignore(numeric_limits::max(),’\n’);

I've found the aforementioned ways but unless there is no other more viable option available I'd rather not use these since it involves including external libraries #limits and #vector receptively.

Aproximation :

• cin.ignore() is the closest and I think it would suffice.

Answer 1

In C scanf("%[^\n]%*c"); would remove all the redundant characters like spaces, newline, alphaNumerics form the buffer before/ after taking another input.

This is amiss in so many ways.

1. "%[^\n]%*c" scans 1 or more non-'\n' (Attempting to save them) and then 1 '\n'. There must exist a leading non-'\n' else the scanning stops. There is nothing special about spaces and alphanumeric - just '\n' and non-'\n'.

2. Undefined behavior: "%[^\n]" lacks a matching pointer to save the input. Even with a matching char *, it lacks a width and is prone to buffer overflow. It is worse than gets().

3. With an input of only "\n", nothing is consumed, nothing is saved. scanf("%[^\n]%*c"); fails to consume anything if the first character is a '\n'. Without checking the return value, the calling code does not know if anything was read. The matching char *, if it was there, is unchanged or potentially indeterminate.

Do not use scanf("%[^\n]%*c"); or its C++ equivalent std::scanf("%[^\n]%*c");

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OP's update to scanf("%d%*[^\n]%*c", &num); is still faulty and should not be used in robust code.

4. It risks undefined behavior due to int overflow. It does not check the return value.

5. It does not consume the rest of the line when the numeric text is followed by a '\n'. That '\n' remains in stdin.

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A better C approach begins with fgets() and checking return values.

Answer 2

Using C++20 onwards, you might want to use std::format. This is basically an implementation of FMT into the standard C++.

For more complex scenarios regular expressions should help.