按 c 中的地址打印和编辑指针

Question

I want to print the value into pointer and edit the pointer's value that located at 0xabcd address

For example 0xabcd ->0xeeeee, I want that 0xabcd will point to 0xaaaa.

int * buff = 0xabcd;
print("the value is %p",*buff); // here is want to see 0xeeeee 
*buff =0xaaaa;

is that right?

Answer 1

No, it is not.

int * buff = 0xabcd;

It assigns the pointer with the integer value converted to pointer to int. It is very very unlikely the address 0xabcd to point to the int array. Dereferencing it invokes Undefined Behaviour.

The %p is to print pointer not the integer referenced by it. Second UB

int a,b;
int * buff = &a;
print("the value is %p",(void *)buff); 
*buff =0xaaaa;
print("the value is %d",*buff); 
buff = &b;
print("the value is %p",(void *)buff); 

But this form us used very often in the low level programming where we know the addresses of the hardware registers or buffers.

Example:

volatile int *ADC_Result = (volatile int *)0x45000000;

Then you can see what is in this register

printf("ADC value: %d\n", *ADC_Result);

But you need to use the correct format (in this case %d to print the integer)