Why Does Printing a Function Name in C Result in \'1\'?

couting a Function Without Calling: Unraveling the Mystery

Imagine this: you're coding away and instead of invoking a function with parentheses, you simply print its name. Surprisingly, the result is always 1. This unconventional approach leaves you perplexed, both about the 1s and the absence of the expected function pointer.

Let's delve into the intricacies of your code:

Contrary to your intention, you're not actually calling the function pr in the cout<< statements. Instead, you're passing its function pointer to cout. This causes pr to be implicitly converted to a Boolean value, resulting in the ubiquitous 1s in your output.

To enhance your understanding, consider using cout << boolalpha before printing. This will display true instead of 1, reinforcing the Boolean nature of the output.

While C 11 offers an elegant solution:

This overload allows you to print function pointers of arbitrary arity, displaying information like the function pointer's address and the number of its arguments.

So, there you have it: the enigmatic 1s are not a mere coincidence but a result of implicit type conversion. Understanding this behavior is crucial for avoiding unexpected outcomes and crafting more robust code.

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