C通过解剖第8章练习6 [英] C By Dissection chapter 8 exercise 6
问题描述
您好，我想知道是否有人可以帮助我以下内容？
编写一个以循环方式移动五个字符变量的存储值的函数。您的功能应该以下列方式工作。假设c1，c2，...，c5是char类型的变量，并假设这些变量的值分别是A，B，......E。函数调用shift（& c1，& c2，& c3，& c4，& c5）应该使变量c1，c2，...，c5具有值''B''''' C''，'D''，''E''，''A''。你的函数定义应该如下开始：
void shift（char * p1，char * p2，char * p3，char * p4，char * p5）
{
....
}
通过五次调用来测试你的功能并依次打印出来BCDEA，CDEAB，DEABC，EABCD和ABCDE。
--------------------------- -------------------------------------------------- -----------------------------------
我在想为什么练习要求启动上面的功能定义？
任何帮助都将不胜感激。
提前谢谢。
Erm，因为有5个字符变量需要更改，因此需要5个输入参数，因为函数必须更改它们传递的变量的值，所以需要通过引用传递（即char *）而不是值（char）。
如果你遇到麻烦帽子问题然后尝试这个更简单的相关问题。
编写一个函数来交换2个字符变量的存储值。假设c1和c2是char类型的变量，并假设这些变量的值分别是''A''和''B''。函数调用swop（& c1，& c2）应该使变量c1和c2分别具有值B和A。你的函数定义应该如下开始：
void swop（char * p1，char * p2）
{
.. ..
}
通过调用两次并依次打印BA和AB来测试你的功能。
< blockquote>对不起，我没跟着你。我有我的程序书，但没有关于切换的例子。我试图想出这个但是...... ????这就是我已经拥有的。
#include< stdio.h>
void shift（char * p1，char * p2，char * p3，char * p4，char * p5）
{
char c1 =''A''，c2 =''B''，c3 =''C''，c4 =''D''，c5 =''E'';
int tmp = 0;
}
我过早地试图编译它并返回错误。
我希望有人能把我放在正确的轨道上。
void shift（char * p1，char * p2，char * p3，char * p4，char * p5）
{
char temp;
temp = * p1; * p1 = * p2; * p2 = * p3; * p3 = * p4; * P4 = P5 *; * p5 = temp;
}
Hello, I was wondering if someone could help me with the following?
Write a function that shifts the stored value of five character variables in a circular fashion. Your function should work in the following way. Suppose that c1, c2,...,c5 are variables of type char, and suppose that the values of these variables are ''A'' , ''B'' , ....'' E'', respectively. The function call shift (&c1, &c2, &c3, &c4, &c5) should cause the variables c1, c2,..., c5 to have the values ''B'', ''C'', ''D'', ''E'', ''A'', respectively. Your function definition should start as follows:
void shift(char *p1, char *p2, char *p3, char *p4, char *p5)
{
....
}
Test your function by calling it five times and printing out, in turn, BCDEA, CDEAB, DEABC, EABCD, and ABCDE.
----------------------------------------------------------------------------------------------------------------
I was wondering why the exercise calls for starting the function definition as above?
Any help would be greatly appreciated.
Thank you in advance.
Erm because there are 5 character variables to be changed so 5 input parameters are required and since the function has to change the value of the variables they are passed they need to be passed by reference (i.e. char *) rather than by value (char).
If you are having trouble getting you head round that problem then try this simpler related problem.
Write a function that swops the stored value of 2 character variables. Suppose that c1 and c2 are variables of type char, and suppose that the values of these variables are ''A'' and ''B'' respectively. The function call swop(&c1, &c2) should cause the variables c1 and c2 to have the values ''B'' and ''A'' respectively. Your function definition should start as follows:
void swop(char *p1, char *p2)
{
....
}
Test your function by calling it twice and printing out, in turn, BA and AB.
I am sorry I am not following you. I have my the program book, but there is no examples that talk about switching. I have tried to figure this out but.....???? Here is what I have already.
#include <stdio.h>
void shift(char *p1, char *p2, char *p3, char *p4, char *p5)
{
char c1 = ''A'', c2 = ''B'', c3 = ''C'', c4 = ''D'', c5 = ''E'';
int tmp = 0;
}
I have prematurely tried to compile this and come back with errors.
I am hoping that someone can put me on the right track.
void shift(char *p1, char *p2, char *p3, char *p4, char *p5)
{
char temp;
temp=*p1; *p1 = *p2; *p2 = *p3; *p3= *p4; *p4=*p5; *p5 = temp;
}
这篇关于C通过解剖第8章练习6的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
