"C language Written review" pointer arrays and arrays pointers

This week began to do C language pen test, what! The test is the concept, a variety of details, although it seems that the book can be found, but I really do not know ... Suddenly there is a kind of my C-language good slag feeling t_t

Well, next to the "on the computer Experiment Blue Book, comprehensive test question two" in the difficult 2 questions, these two questions make you to the "pointer array" and "array pointer" the difference is clearer.

"Example 1"

The following procedures are available:

1#include <stdio.h>2 3 intMain ()4 {5     Char*s[] = {" One"," Both","three"}, *p;6p = s[1];7printf"%c,%s\n", * (p+1), s[0]);8     return 0;9}

The result after execution is _______.

A. N, B. W, one C. T, one D. O, and

The first time I chose C, because I thought (p+1) is pointing to "three". This is an understanding error for the pointer p type.

The correct answer is B, because:

1#include <stdio.h>2 3 intMain ()4 {5     Char*s[] = {" One"," Both","three"};//S is an array of pointers, the element is 36                                          //pointer to a string constant7     Char*p = s[1];//p is a pointer variable that points to a string8printf"%c,%s\n", * (p+1), s[0]);9     //(p+1) is the address of P plus the size of a character memory, from point T to point WTen     return 0; One}

So, usually we say that the pointer to the string, in fact, is pointing to a character, so it is a displacement operation, the addition and subtraction are 1.

Also, if you put the 8th line

    printf ("%c,%s\n", * (p+1), s[0]);

Change into　　　

    printf ("%c,%s\n", * (p+1), s[0]);

The output will be: Wo,one

Because when we output a string, we actually pass the first address of the string to the printf () function, which is judged by the end of the ' \ n '.

"Example 2"

1#include <stdio.h>2 3 intMain ()4 {5     inta[3][4] = {{1,2,3,4}, {5,6,7,8}, {9,Ten, One, A}};6     int(*p) [4] =A;7printf"%d\n", * (* (p+1)+3));8     return 0;9}

The result of the above code execution is _______.

The answer is 8.

I don't understand the problem at first. How can * also *, (*P) [4] What Ghost ... I messed up the "pointer array" and "array pointers" before I made this paper =

That is true:

1. (*p) [4]: declares that P is a pointer to (an array of 4 int elements), so p+1 from Point A[0] to a[1]

2.* (* (p+1) +3): Why are there two stars?

* (p+1) = a[1][0] (i.e. 5), the first star from a[1] to a[1][0], although the address has not changed, but the type of pointer changed! The original pointer +1 is a plus 4 int, now the pointer + 1 only add an int!!

* (p+1) +3 = A[1][3] (that is, 8), then take a * is to read from the address 8

You can also modify the original code to clarify:

1#include <stdio.h>2 3 intMain ()4 {5     inta[3][4] = {{1,2,3,4}, {5,6,7,8}, {9,Ten, One, A}};6     int(*p) [4] =A; 7printf"P:%p\n", p);8printf"p+1:%p\n", p+1);9printf"* (p+1):%p\n", * (p+1));Tenprintf"* (p+1) + 3:%p\n", * (p+1)+3); Oneprintf"%d\n", * (* (p+1)+3)); A     return 0; -}

The result of the output is:

0x7fff9f4ba180 P+10x7fff9f4ba190* (p+10x7fff9f4ba190* (p+1) +3 0x7fff9f4ba19c 8

Oh, yes.

"C language Written review" pointer arrays and arrays pointers