C 逗号在 scanf 中破坏了 integer

Question

The code below asks the user to enter a number followed by an operator. It works as long as no commas are used. For example,

Happy path

6000+2000 will output 6000.000 + 2000.000 

Unexpected output

6,000+2,000 will output 6.000 , 0.000

Can scanf somehow treat the value as float when using a comma?

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

int main(int argc, char *argv[]) 
{
    char operatorFound;
    float num1,num2, calcValue;
    printf("Please enter a simple expression (ie: 2+2):" );
    scanf("%f%c%f", &num1, &operatorFound, &num2);
    printf("%f\n%c\n%f", num1, operatorFound, num2); 
   return 0;
}

Answer 1

It may be possible to adjust scanf under Linux by setting the locale accordingly. See scanf Linux documentation

For decimal conversions, an optional quote character ('). This specifies that the input number may include thousands' separators as defined by the LC_NUMERIC category of the current locale. (See setlocale(3).) The quote character may precede or follow the '*' assignment- suppression character.

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

int main(int argc, char *argv[]) 
{
    char operatorFound;
    float num1, num2, calcValue;
    // Use en_US.UTF-8 to separate groups of thousands, many other countries use a period instead, and some countries separate thousands groups with a space
    setlocale(LC_NUMERIC, "en_US.UTF-8");
    puts("Please enter a simple expression (ie: 2+2):" );
    
    // Notice the use of a single quote
    if (scanf("%'f %c%'f", &num1, &operatorFound, &num2) == 3) { 
        printf("%f\n%c\n%f", num1, operatorFound, num2); 
    } else {
        puts("Unable to process input");
    }
   return 0;
}

Answer 2

While you are scanning the input C takes commas as char.

If you enter only 6,00 it will still print 6.000, 0.000 because it takes comma as a second input.

Therefore you should use dots '.' for floats

Answer 3

So after testing. I found out my compiler is flawed. When I test on Linux GCC all the answers worked.

If I enter

6,000 + 2,000 it will output 8000

6000 + 2000 it will output 8000

6,000 + 2000 it will output 8000

6000 + 2,000 it will output 8000

6000+2000 it will output 8000

etc etc

Thank you all for looking into this and helping me out