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Possible Duplicate:
Behaviour of printf when printing a %d without supplying variable name

What happens if I use, for example, printf("%d %d"); ? Will it just pop the last eight bytes from the stack and print them out?

Community
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Zakum
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4 Answers4

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In GCC - you get a warning (this is done using __attribute__ ((__warn_unused_result__))).

On x86 you don't get a stack error, as the caller will push the data to the stack, and also pop after the function returned. This is called the C calling convention, unlike pascal - in which the function will also pop the data from the stack (using ret 10 in ASM for example).

The values of the data you required will be random.

BenMorel
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elcuco
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Technically its undefined behaviour if the number of format specifiers in printf() is greater than the number of arguments.

However the following is fine

printf("%d",x,y); // y is evaluated but not printed.
Prasoon Saurav
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In that case, you'll get garbage data which relies upon the compiler and its compiling option....

Akash KC
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It will print garbage values but it has 'More % conversions than data arguments' warning.

Prasad G
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