Is this printf output for anything?

Hallo,
I have this this example, but is this usefull in practical use?
Normaly I would work with IF or CASE, but this version can only answer 0 or 1.
But can I use it an put the result in a variable or something with this system?
Orf write different text statements?

Code:

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#include <stdio.h>
#include <conio.h>
int main(void){
    int temper = 20;
    printf("Ist kleiner als 23°C %d\n", temper < 23);
    printf("Ist größer als 23°C %d\n", temper  > 23);
    return 0;   
}

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HM, ok, this doesn't work
I hoped it is like an IF oder Case

Code:

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    c=5 == 8 > 6;
    printf("C ist %d\n", c);

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But why is always C=0?

Code:

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        int c=99;




        c=5 == 6 > 1;
        printf("C ist %d\n", c);

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The result is always 0?! Why not 99?

Comarison operations yield an int with 0 representing logical false, and 1 (or any other value) representing logical true. You have to specify string literals that you conditionally pass to printf, along with the %s format specifier for a string. Like

Code:

---

#include <stdio.h>

int main(void)
{
  static const char eq[] = "gleich";
  static const char lt[] = "kleiner als";
  static const char gt[] = "größer als";


  int temper = 25;
  printf("Ist %s 23°C.\n", temper == 23 ? eq : (temper < 23 ? lt : gt));
}

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I used inlined ternary operators here. You could also use if-else or switch-case to assign the string to a variable first.