C语言实现简单计算器功能V1.1

#define _CRT_SECURE_NO_WARNINGS #pragma warning(disable:4996) //以上两行避免strcat与sprintf报错 #include <stdio.h> #include <math.h> #include <string.h> #include <Windows.h>//避免 C4013 “_sleep”未定义；假设外部返回 int 警告 #include "unistd.h"//引用内部头文件unistd.h用于延时 void OutputThePanel(void); void swap(float* a, float*b); int main(void) { int Oper;//存储操作序号 float a, b;//存储需要计算的数据 float Num[50];//存储9号功能的数据，也就是排序的初始数据 char HistoryOperate[5][1250] = { "EMPTY", "EMPTY", "EMPTY", "EMPTY", "EMPTY" };//用于存储五步历史步骤，0表示最新的记录 /*当历史记录为"EMPTY"时，表示这个位置目前是空的*/ int i, j;//循环参数 double Ans;//用于存储答案 char A[16] = "", B[16] = "", ANS[16] = "";//用于储存输入的数据与答案的字符串形式 do { OutputThePanel(); //输出面板 OperateWrong://如果输入的编号不是1-9的整数，就跳回这里让用户重新输入 printf_s("\n\033[35m*\033[32mPlease input the number of the operate you want:\033[0m"); //以上代码负责显示界面 /* \033[32m表示绿色输出，\033[0m表示白色输出 \033[35m表示紫色输出，\033[31m表示红色输出 \033[33m表示黄色输出 */ scanf_s("%d", &Oper); if ((Oper != 1) && (Oper != 2) && (Oper != 3) && (Oper != 4) && (Oper != 5) && (Oper != 6) && (Oper != 7) && (Oper != 8) && (Oper != 9) && (Oper != 10) && (Oper != 11) && (Oper != 12) && (Oper != 13) && (Oper != 14))//判断是否符合输入操作编号为1-9的整数 { printf_s("[\033[31mERROR\033[0m]\033[60mThe Number You Input Is Not Allowed!\033[0m\n"); printf_s("Please try it again.Choose your operate and input the number of it.\n"); goto OperateWrong;//不符合输入条件则警告后跳回输入步骤 } //以上确保输入的操作是符合规定的操作序号，也就是一定符合1<=Oper<=9且是整数 switch (Oper) { case 1://当选择1操作时应当为加法操作 printf_s("\n\033[35m*\033[32mPlease input the two numbers you want to sum up:\033[0m"); //提示用户输入两个需要相加的数字 scanf_s("%f", &a); scanf_s("%f", &b);//读取两个float浮点数 Ans = a + b; for (i = 3; i >= 0; i--)//用原来的0-3总计四条记录覆盖掉1-4的记录，实现记录的后移，为新记录腾出空间 { strcpy(HistoryOperate[i + 1], HistoryOperate[i]); } if ((ceil(a) == (int)a) && (floor(a) == (int)a)) sprintf(A, "%d", (int)a); else sprintf(A, "%0.4f", a); if ((ceil(b) == (int)b) && (floor(b) == (int)b)) sprintf(B, "%d", (int)b); else sprintf(B, "%0.4f", b); if ((ceil(Ans) == (int)Ans) && (floor(Ans) == (int)Ans)) sprintf(ANS, "%d", (int)Ans); else sprintf(ANS, "%0.4f", Ans); //将a、b、Ans从float型数据转为字符串并分别存储与A、B、ANS中 strcpy(HistoryOperate[0], "");//清空 strcat(HistoryOperate[0], "\033[32m "); strcat(HistoryOperate[0], A); strcat(HistoryOperate[0], " + "); strcat(HistoryOperate[0], B); strcat(HistoryOperate[0], " = "); strcat(HistoryOperate[0], ANS); strcat(HistoryOperate[0], "\033[0m"); //连接字符串函数，用于存储最新的记录 //分多句连接避免"strcat实参过多"警告 //存储记录时就把颜色存进去，增强页面美化的兼容性 printf_s("\033[35m>\033[32mThe answer is :\033[0m%s\n", ANS); printf_s("\n\033[35m*\033[32mThe operate has been saved temporarily.\n\033[35m*\033[32mIf you exit the software,the memory will lost.\033[0m"); break; case 2://当选择1操作时应当为减法操作 printf_s("\n\033[35m*\033[32mPlease input the two numbers you want to subtract:\033[0m"); //提示用户输入两个需要相减的数字 scanf_s("%f", &a); scanf_s("%f", &b);//读取两个float浮点数 Ans = a - b; for (i = 3; i >= 0; i--)//用原来的0-3总计四条记录覆盖掉1-4的记录，实现记录的后移，为新记录腾出空间 { strcpy(HistoryOperate[i + 1], HistoryOperate[i]); } if ((ceil(a) == (int)a) && (floor(a) == (int)a)) sprintf(A, "%d", (int)a); else sprintf(A, "%0.4f", a); if ((ceil(b) == (int)b) && (floor(b) == (int)b)) sprintf(B, "%d", (int)b); else sprintf(B, "%0.4f", b); if ((ceil(Ans) == (int)Ans) && (floor(Ans) == (int)Ans)) sprintf(ANS, "%d", (int)Ans); else sprintf(ANS, "%0.4f", Ans); //将a、b、Ans从float型数据转为字符串并分别存储与A、B、ANS中 strcpy(HistoryOperate[0], "");//清空 strcat(HistoryOperate[0], "\033[32m "); strcat(HistoryOperate[0], A); strcat(HistoryOperate[0], " - "); strcat(HistoryOperate[0], B); strcat(HistoryOperate[0], " = "); strcat(HistoryOperate[0], ANS); strcat(HistoryOperate[0], "\033[0m"); //连接字符串函数，用于存储最新的记录 //分多句连接避免"strcat实参过多"警告 //存储记录时就把颜色存进去，增强页面美化的兼容性 printf_s("\033[35m>\033[32mThe answer is :\033[0m%s\n", ANS); printf_s("\n\033[35m*\033[32mThe operate has been saved temporarily.\n\033[35m*\033[32mIf you exit the software,the memory will lost.\033[0m"); break; case 3://当选择1操作时应当为乘法操作 printf_s("\n\033[35m*\033[32mPlease input the two numbers you want to multiply:\033[0m"); //提示用户输入两个需要相乘的数字 scanf_s("%f", &a); scanf_s("%f", &b);//读取两个float浮点数 Ans = a * b; for (i = 3; i >= 0; i--)//用原来的0-3总计四条记录覆盖掉1-4的记录，实现记录的后移，为新记录腾出空间 { strcpy(HistoryOperate[i + 1], HistoryOperate[i]); } if ((ceil(a) == (int)a) && (floor(a) == (int)a)) sprintf(A, "%d", (int)a); else sprintf(A, "%0.4f", a); if ((ceil(b) == (int)b) && (floor(b) == (int)b)) sprintf(B, "%d", (int)b); else sprintf(B, "%0.4f", b); if ((ceil(Ans) == (int)Ans) && (floor(Ans) == (int)Ans)) sprintf(ANS, "%d", (int)Ans); else sprintf(ANS, "%0.4f", Ans); //将a、b、Ans从float型数据转为字符串并分别存储与A、B、ANS中 strcpy(HistoryOperate[0], "");//清空 strcat(HistoryOperate[0], "\033[32m "); strcat(HistoryOperate[0], A); strcat(HistoryOperate[0], " * "); strcat(HistoryOperate[0], B); strcat(HistoryOperate[0], " = "); strcat(HistoryOperate[0], ANS); strcat(HistoryOperate[0], "\033[0m"); //连接字符串函数，用于存储最新的记录 //分多句连接避免"strcat实参过多"警告 //存储记录时就把颜色存进去，增强页面美化的兼容性 printf_s("\033[35m>\033[32mThe answer is :\033[0m%s\n", ANS); printf_s("\n\033[35m*\033[32mThe operate has been saved temporarily.\n\033[35m*\033[32mIf you exit the software,the memory will lost.\033[0m"); break; case 4://当选择1操作时应当为除法操作 DividedByZero://如果输入分母为0，则重新询问数据 printf_s("\n\033[35m*\033[32mPlease input the two numbers you want to divide:\033[0m"); //提示用户输入两个需要相除的数字 scanf_s("%f", &a); scanf_s("%f", &b);//读取两个float浮点数 if (b != 0)//分母不能为0 { Ans = a / b; for (i = 3; i >= 0; i--)//用原来的0-3总计四条记录覆盖掉1-4的记录，实现记录的后移，为新记录腾出空间 { strcpy(HistoryOperate[i + 1], HistoryOperate[i]); } if ((ceil(a) == (int)a) && (floor(a) == (int)a)) sprintf(A, "%d", (int)a); else sprintf(A, "%0.4f", a); if ((ceil(b) == (int)b) && (floor(b) == (int)b)) sprintf(B, "%d", (int)b); else sprintf(B, "%0.4f", b);