第4堂课后作业

作业1：

Visual Studio 2012 主要用于C++、C#和VB语言的开发

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作业2：

1，分析问题

根据问题来设计一种解决方案。

2编制程序

通过程序语言严格描述这个方案。

3编译

编译中发现错误，要分析原因，修改源程序。

4连接

连接中发生错误，要分析原因，修改源程序。

5调试运行

调试运行中发现问题分析本身有错误，要重新分析问题.

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作业3-1：

1) -abc    8) #micro   合法的变量名，在c语言中，只能由字母，数字和下滑线组成，且第一位只能是字母或下划线。

3) for    11) while       c语言中的基本字不能成为变量名

作业3-2:

(1):int的特点是保存整数，常用于年龄、月份等数据的保存。
(2):无符号整型unsigned int 和int类似，但只取正值。

(3):短整型 short int 在32位机中占2个字节

(4):长整型 long int 在32位机中占8个字节

(5): 无符号的长整型的最高位不代表正负，所以它的大小比普通长整大，但同时也失去了表示负数的功能

(6): 字符型 char 表示字符串
(7):无符号字符型 unsigned char   unsigned是无符号字符，只有正没有负，char是8位，是0~255

(8) :单精度 float   float的精度是6位有效数字，取值范围是10的-38次方到10的38次方，float占用4字节空间
(9): 双精度double   double的精度是15位有效数字，取值范围是10的-308次方到10的308次方,double占用8字节空间
(10) 长双精度 long double   long double的精度不少于double的精度，就像int和long int一样

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作业4：

#include<stdio.h>
void main()
{
 char tip[11]="yanhaofeng";
  printf("%c %c\n",tip [0],tip[3]);
}
效果图

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作业5：

#include <stdio.h>

int main()

{

printf("%d\n",139133);

printf("%f\n",3.1415926);

}

作业6

6-1

(1):#include<stdio.h>
void main()
{
 int x;
 x= 25 + 0125;
printf("x=%d",x);
}

(2)#include<stdio.h>
void main()
{
 int x;
 x=24 * 3 / 5 + 6;
printf("x=%d\n",x);
}

(3)#include<stdio.h>
void main()
{
 int x;
 x=  36 + - (5 - 23 ) / 4;
printf("x=%d\n",x);
}

(4)#include<stdio.h>
void main()
{
 int x;
 x= 35 * 012 + 27 / 4 / 7 * (12 - 4);
printf("x=%d\n",x);
}

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6-2

(1)#include<stdio.h>
#include<math.h>
void main()
{
 int x;
 x=3 * (2L + 4.5f) - 012 + 44 ;
printf("x=%d\n",x);
}

(2)#include<stdio.h>
#include<math.h>
void main()
{
 int x;
 x=3 * (int)sqrt(144.0);
printf("x=%d\n",x);
}

(3)#include<stdio.h>
#include<math.h>
void main()
{
 int x;
 x=cos(2.5f + 4) - 6 *27L + 1526 - 2.4L ;
printf("x=%d\n",x);
}

 

由浮点型转换为整型

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