如果我在 scanf 中传递变量值而不是变量引用会发生什么？

Question

I have this code:

int a;

scanf("%d", &a);

printf("You entered: %d", a);

which works perfectly fine, but I wondered what would happen if I pass the variable value instead of the variable reference in scanf like scanf("%d", a) .

I don't get an error from the compiler but there is no output either. What is happening here?

Answer 1

In scanf "%d" is expecting a pointer to int meaning a pointer to the address of the variable where the value will be stored, if you pass a you will just be passing int there is no way the value can be stored because scanf won't know where.

Doing this will invoke undefined behavior .

Answer 2

Basically the & is an operator that returns the address of the operand. If you give in scanf the variable a without the &, it will be passed to it by-value, which means scanf will not be able to set its value for you to see. Passing it by-reference - using & actually passes a pointer to a - allows scanf to set it.

Answer 3

scanf expects the argument corresponding to the %d conversion specifier to be the address of an int object (with type int * ). Whatever value a contains will be interpreted as an address ¹ .

The behavior on doing this is undefined - the compiler isn't required to issue a diagnostic during translation, and the runtime environment isn't required to throw any kind of exception. Depending on the value contained in a you may get a runtime error, or you may clobber another data object, or your code may work without any apparent errors, or something completely different may happen.

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1. At least on systems where sizeof (int) == sizeof (int *) - for a system like x86_64, that's not the case, so scanf will treat bytes outside of a as part of an address.