为什么参数太少的 printf 在 Cygwin 上工作？

Question

int main()
int a = 1 , b=2 ,c=3 ,d=4 ;
a= ++b;
c= d++;
print f("a = %d , b = %d , c = %d, d= %d" , a , b ,c);
return 0;}
a=3 b=3 c=4 d=5

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I don't know why works this code nomally in cygwin because I dont write about D in printf

Answer 1

You are relying on undefined behavior by calling printf with too few arguments. You should never do this as the program will likely crash, and if it does run, the results will be randomly based on the specific compiler/system.

What is happening here is that printf is popping more variables from the stack than it is supposed too and is grabbing something from the stack from main . This is almost always going to cause some sort of stack/memory corruption later on in your program. Of course your code is short and so this is not happening in this simple case.