Questions about C pointers, C pointers

Scope. You static char *chh;

static char *ch1;Although the address pointed to by the two pointers does not change, have you ever wondered whether the memory address they point to has been released, char chc[10]; It's local, the function is out, the lifecycle is over, and you're trying to access it with a pointer in void times().

People are already hanging, and you still want to dig up the whip corpse, you are too ruthless.

There are other ways to do this, but I recommend that you change that first.

char chc[10];, change it to a global variable, and adjust it yourself.

Wow that's a long time, what's the problem? You don't say anything, how do I answer it?

Describe the problem clearly, is there a bug in this program, or you can't get the desired result、、、

Static: The address will not change, it will always be in memory until the program is terminated.

static changes the duration of local variables.

For example, you here, after the first execution.

CHH is 15 (for example).

Until the next execution, *chh is still 15

Static changes the visible range of global variables, originally the global variable is accessible to all programs, if static is added, it can only be accessed by this subprogram.

As in the case by more than one. c(.cpp) file, in order to prevent other source files, a global variable with the same name is defined, and a static is added in front of it to solve the problem.

int a[2][3], p[3]；

Analysis: For a, there is no doubt that a is a two-dimensional array, and the value of a zhi points to the origin dao address of the int type. Inside.

For p, first p comes first'[ ]'Combined with p[3], it means that p-is an array, and p is then with'＊'Combined' *p[3] ', which represents the time pointer variable stored in the p array, and finally combines it with the int type' int *p[3] ', which indicates that the array is stored with a pointer of type int.

p[0]=&a[1][2];

So the value of p[0] is a pointer to type int.

The value of a[1][2] is an int type, plus'&'The result is a pointer to the int type.

So p[0]=&a[1][2]; Establish.

It is recommended that ZL check out "Let You No Longer Be Afraid of the Pointer".

*p[3] is an array of pointers, and each array element is an int pointer.

The mistake with a is that p is the pointer to this pointer array, and if you change the value of p, then this pointer array will become garbage.

*p[3] This is the pointer array, and the elements inside it are the pointers.

Normally, A is fine, but it may be an error. A forced turn is required, but addressing is fine.

If there is a definition statement: zhiint a[2][3], *p[3]; Then the correct DAO in the following statement is (c).

a. p=a；Error, anachronic answer mismatch.

Error, type mismatch.

That's right. Error, type mismatch.

The test procedure is as follows. The following program can be compiled successfully, and replacing it with something else will not succeed.

#include

void main()