How can I use variable width specifiers defined by a macro in sscanf()

Question

I am trying to specify the width of variables in a sscanf() call based on a #define that calculates its values based on another #define.

It works when the TEST define is just a single number, but when it becomes a calculation, the code fails. Have TEST call another function that contains the calculation does not work either.

The code I am working with:

#include <stdio.h>

#define A       3
#define TEST    FUN()
#define FUN()   (A + 2)
#define STR(X)  _STR(X)
#define _STR(x) #x

int main()
{
    char input[] = "Test123";
    char output[10];

    sscanf(input, "%" STR(TEST) "s\n", output);

    printf("%s\n", output);

    return 0;
}

What am I missing here?

The preprocessor doesn't to any arithmetic, it's basically a plain search and replace. — Some programmer dude, Nov 11 '19 at 12:45
I think that `input` and `output` are swapped in the code. Destination string, in scanf, is the first parameter. — Roberto Caboni, Nov 11 '19 at 12:46
1) Does it file in compilation phase or at runtime? 2) What output do you get? — Roberto Caboni, Nov 11 '19 at 12:47
You might consider _variable_ as something you pass in, either via command line parameter or as a variable argument to a function. But macros are pre-processed, and not variable during runtime. — ryyker, Nov 11 '19 at 12:48

Soner from The Ottoman Empire · Answer 1 · 2019-11-14T06:06:16.483

3

You cannot realize an arithmetic operation as happened in your question. So, it is evaluated as (3 + 2) in string form while compiling.

My two cents reminiscent of the post,

#define STRING_LENGTH 20
#define STR(x) _STR(x)
#define _STR(x) #x

{
  ...    
  fscanf(input, "%" STR(STRING_LENGTH) "s", output);
  ...
}

Two macros are required because if you tried to use something like _STR directly you would just get the string "STRING_LENGTH" instead of its value.

For more click here and an example.

edited Nov 14 '19 at 06:06

answered Nov 11 '19 at 12:48

Soner from The Ottoman Empire

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Thank you. Are there any good alternatives for what I am trying to achieve? – Kodiak Nov 11 '19 at 12:50

KamilCuk · Accepted Answer · 2019-11-11T13:08:44.830

Just build the scanf format string during execution. For example:

const int width_modifier = FUN();
char fmt[100] = {0};
snprintf(fmt, sizeof(fmt), "%%%ds\n", width_modifier);
sscanf(input, fmt, output);

What am I missing here?

Preprocessor does text replacement, it doesn't calculate things. So "%" STR(TEST) "s\n" expands to a string containing "%" "3 + 2" "s\n", which after concatenation is "%3 + 2s", which is an invalid scanf format string.

Alternatively you can use other program to prepare the source file for compilation. A popular use is a preprocessor that preprocesses the file before compilation and that is able to calculate arithmetic. A popular choice is m4:

m4_define(`A', 3)
m4_define(`FUN', `eval(A + 2)')

#include <stdio.h>
#define STR(X)  #X
int main() {
    printf("%s", STR(FUN()));
}

preprocessed with m4 and compiled would output 5.

score 0 · Answer 3 · answered Jul 09 '22 at 01:05

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


#define MAX_LENGTH 5


int main()
{
    // specify format 

    char formatStr[20] = {0};
    snprintf(formatStr, 20-1, "%s%d%s", "%", MAX_LENGTH-1, "s%u");
    printf("%s\n", formatStr);
    
    // example to use that format 

    char *input = "programming 123";
    
    char dataStr[MAX_LENGTH] = {0};
    int  dataVal;
    sscanf(input, formatStr, dataStr, &dataVal);
    printf("%s-%d\n", dataStr, dataVal);
    
    return 0;
}

output:

%4s%u
prog-21942

How can I use variable width specifiers defined by a macro in sscanf()

3 Answers3

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