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:
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