我正在尝试了解特定的函数指针和赋值

Question

I am reading the source code of the video game “Rise of the Triad: Dark War”, and I came across a line of code I don't understand.

void (*USL_MeasureString)(char *, int *, int *, font_t *) = (void (*)(char *, int *, int *, font_t *))VW_MeasurePropString, (*USL_DrawString)(char *) = VWB_DrawPropString;

As far as I understand it void (*USL_MeasureString)(char *, int *, int *, font_t *) means *USL_MeasureString is a function pointer that takes char *, int *, int *, font_t * as parameters and returns nothing because of void .

But then, I am confused about the rest of the code.

What does it means exactly?

Best regards.

Answer 1

Let's break this long line into three parts:

void
  (*USL_MeasureString)(char *, int *, int *, font_t *) = (void (*)(char *, int *, int *, font_t *))VW_MeasurePropString,
  (*USL_DrawString)(char *) = VWB_DrawPropString;

I believe it's easier to understand now. As you've correctly figured out, USL_MeasureString is defined as a function pointer, and the RHS (Right-Hand Side) of the assignment is a type cast applied to the word VW_MeasurePropString .

The second part appears tricky on its own, but once you get to know how the following line works, you'll know it's the same:

char (*a) = (char *)"123", (*b) = "xyz";

Here, both a and b are pointers to char , so you can think of char as a "base type" for the whole line of variable definitions. And of course you can rewrite it as:

char (*a) = (char *)"123";
char (*b) = "xyz";

So the part after the comma in your original code can be rewritten as (where void is the "base type"):

void (*USL_DrawString)(char *) = VWB_DrawPropString;

As you can see, it's just the definition of another function pointer with a different argument list and no type casting.