How does printf in c++ process the arguments that are passed to it? -

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this question has answer here:

i have been trying understand how printf processes arguments passed it. more specific can please explain how following outputs occour.

int a=1; printf("%d %d %d",++a,a,a++);// outputs: 3 3 1  a=1; printf("%d %d %d",a++,a,a++);// outputs: 2 3 1  a=1; printf("%d %d %d",a,a,a++);// outputs: 2 2 1  a=1; printf("%d %d %d",a,a++,a);// outputs: 2 1 2  a=1;     printf("%d %d %d",a,a,++a);// outputs: 2 2 2

the same output occurs cout statement.

this code

printf("%d %d %d",++a,a,a++);// outputs: 3 3 1

modifies twice in same expression. precise modifies twice without intervening sequence point. because of this code has undefined behaviour , attempt reason futile. compiler can wants, , different compilers different things. see undefined behavior , sequence points detailed explanation.

with code this

printf("%d %d %d",a,a,a++);// outputs: 2 2 1

it's undefined when a++ incremented, before or after used other arguments. output vary.

the real thing understand here don't write code this. follow rule , won't go wrong.


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