This C program outputs garbage [closed]

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#include<stdio.h>
int main()
{
    char a='x'; 
    printf("%c %d",a);
    return 0;
} 

Output:

x 134513696

What is 134513696?

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Garbage. This is due to a programming error: You put only one parameter on the stack (a), but printf takes 2 values from the stack, because of the two percent signs.

If you intended to have both outputs, the character and its ordinal value, you should have written printf("%c %d", a, a);

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Suppose you enter into a contract with a mafia boss to buy a shipment of goods for $1000. Then instead you only hand over $500, and that night you go home to find a dead kitten in your bed. What did you expect?! C is the mafia boss and you you broke your contract with him. Be glad it was just a useless number on your terminal and not your computer blowing up.

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If the number of format specifiers in printf() is greater than the number of arguments passed the behaviour is undefined.

For example :

printf("%d %d %d", 1, 2); // UB
printf("%f %d %d"); // UB

However if the arguments are greater in number(than the format specifiers) the extra ones are just evaluated and ignored.

For example :

printf("%d" ,1,2); //fine. Prints 1

---

printf uses a variable argument list. It cannot check if the number of arguments in your format string (in your case 2 "%c %d") are equivanlent to the number of arguments inside the va_list (you have just one), so it will grab some undefined value. The compiler will not check for you, if the format string is properly formated.