What does *a in the vodi function (u32 *a) mean to input into the function?

wrdt24435

What does *a in the vodi function (u32 *a) mean to input into the function? [Copy link]

I would like to ask you what does the input of this function mean? Does it mean *p of type u32? In the routine, the input is u32 a[n]; the array is defined like this, and when using this function, it is input as function(a) like this. In addition, in the function, *a++ means pointer address + 1? The input in the routine is the above array, and *a++ means the value of a[x+1] when the routine is executed.

swordroo

The description is a bit hard to understand... What do you mean by void fn(u32 *a); What data is a in the data? If so, then a must be a pointer address of type u32, such as: u32 val = 0; fn( &val); *a++, the value operator is at the same level as the increment operator, and it combines from right to left, so increment first and then value; "*a++ in the routine execution means the value of a[x+1]" I understand what you mean, it should be correct!

star_66666

It's very simple

le062

Go read "C and Pointers" to ensure that you will solve the problem once and for all

姚星宇

void func(u32* a){ } This is easier to understand. What is sent to the function is a pointer, which is a pointer to the u32 type.