Why is the following C code giving such output?

Question

I ran the following code in codeblocks and got the output:
10 20
10 20

int main()
{
    int i=10,j=20;
    printf("%d %d\n",i,j);
    printf("%d %d",i);

    return 0;
}

What is the reason of second 20 ?

Answer 1

Since you're calling printf a second time with no intervening code, the value of j is still on the stack, left from the previous call.

Of course, you shouldn't depend on this behavior. Just because you don't see the bug doesn't mean it's not there. :-)

Answer 2

超出vararg列表末尾的读取是未定义的行为，因此第二个20的原因纯属偶然：您的代码可以打印任何整数，甚至崩溃。