C Primer Plus第六版第六章复习题

6.15复习题

/*
此题中文翻译有误，英文原文
Find the value of quack after each line;
 each of the final five statements uses the value of quack produced by the preceding statement. 
后一项语句使用的是先前语句中生成的值
*/

int quack = 2;
quack += 5; 	执行后quack = 7
quack *= 10;	执行后quack = 70
quack -= 6;	执行后quack = 64
quack /= 8;	执行后quack = 8 quack是int类型0.25截断为0
quack %= 3; 	执行后quack = 0

//如果value 是 int类型 输出如下
 36 18  9  4  2  1
 //如果value是double类型 输出比较奇怪

用%3d打印出来如图：在这里插入图片描述
用%3f打印的结果如图：

直到浮点数下溢为0为止,再者double使用%3d格式说明打印也不对呀

//a.x大于5
	x > 5
//b.scanf（）读取一个名为x的double类型值且失败
	scanf("%lf", &x) != 1
//c.x的值等于5
	x == 5

//a.scanf（）成功读取一个整数
	scanf("%d", &x) == 1
//x不等于5
	x != 5
//x大于或等于20	
	x >= 20

修改如下：

#include <stdio.h>
int main(void)
{
	int i, j, list [10];		//修改后

	for (i = 1: i < 10; i++)	//修改后
	{
		list[i] = 2 * i + 3;
		for (j = 1 ;  j <= i ;  j++)//修改后
			printf(" %d", list[j]);
		printf("\n");
	}				//修改后
}

// 打印 4x8的 $矩阵
#include <stdio.h>
int main(void)
{
	for (int i = 0; i < 4; i++)
	{
		for(int j = 0; j < 8; j++)
			printf("$");
		printf("\n");
	}
	return 0;
}

a.Hi! Hi! Hi! Bye! Bye! Bye! Bye!
b.ACGM

a.	Go west, youn
b.	Hp!xftu-!zpvo
c.	Go west, young
d.	$o west, youn

31|32|33|30|31|32|33|
***
1
5
9
13

2 6
4 8
8 10

***
======
=====
====
===
==

min
10
double
ii.正确

//修改后的程序如下：
#include <stdio.h>
#define SIZE 8
int main(void)
{	
	int by_twos[SIZE];
	int index;

	for (index = 1; index < SIZE; index++)
		by_twos[index] = 2 * index;
	for (index = 1; index < SIZE; index++)
	{	
		printf("%d ", by_twos[index]);
		printf("\n");
	}
	
	return 0;
}

long Functionname(formal parameter)
{
	return long variable;
}

long Func(int parameter)
{
	return (long)parameter * parameter; 
}

1: Hi!
k = 1
k is 1 in the loop
Now k is 3
k = 3
k is 3 in the loop
Now k is 5
k = 5
k is 5 in the loop
Now k is 7
k = 7

C Primer Plus第六版第六章复习题

6.15复习题

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