普通計算器的實現
阿新 • • 發佈:2017-05-20
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1 #include<stdio.h> 2 #include<conio.h> 3 #include<stdlib.h> 4 float get_answer()/*定義求結果的函數*/ 5 { 6 float a,b,answer; 7 char flag; 8 scanf("%f%c%f",&a,&flag,&b); 9 switch(flag) 10 { 11 case ‘+‘: answer=a+b;break; 12 case ‘-‘: answer=a-b;break; 13 case ‘*‘: answer=a*b;break; 14 case ‘/‘: answer=a/b;break; 15 } 16 return answer; 17 } 18 void main()/*主函數*/ 19 { 20 float b,answer; 21 char flag; 22 int point; 23 system("title 普通計算器"); 24 system("color f0"); 25 printf("輸入要保留的小數位:"); 26 scanf("%d",&point);27 printf("就緒!過程中按“Esc”退出!\n"); 28 answer=get_answer(); 29 printf("=%.*f\n",point,answer); 30 while(1) 31 { 32 flag=getch();/*檢測將要執行的命令*/ 33 if(flag==42||flag==43||flag==45||flag==47)/*判斷是否是“+,-,*,/”*/ 34 {printf("%c",flag);/*輸出運算符*/ 35 scanf("%f",&b);36 switch(flag) 37 { 38 case ‘+‘: answer+=b;break; 39 case ‘-‘: answer-=b;break; 40 case ‘*‘: answer*=b;break; 41 case ‘/‘: answer/=b;break; 42 } 43 printf("%.*f",point,answer);} 44 else if(flag==27)/*按Esc退出*/ 45 break; 46 else/*再次執行運算*/ 47 { 48 printf("\n請輸入“a+、-、*、/、b”:\n"); 49 get_answer(); 50 printf("=%.*f\n",point,answer); 51 } 52 } 53 }
普通計算器的實現