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  1. #include <stdio.h>
  2. #include <stdlib.h>
  3.  
  4. /* run this program using the console pauser or add your own getch, system("pause") or input loop */
  5.  
  6. char data[5];//運算元 1234.......... 
  7. char opstack[5];//放運算子 
  8. int top=-1,optop=-1;//top紀錄運算元堆疊、optop紀錄運算子堆疊 
  9.  
  10. char oppop(){
  11. 	char sym;
  12. 	sym=opstack[optop];
  13. 	optop--;
  14. 	return sym;
  15. }
  16.  
  17. void oppush(char strstack){
  18. 	opstack[++optop]=strstack;
  19. }
  20. //上方是運算子、下方是運算元 
  21. int pop(){
  22. 	int num;
  23. 	num=data[top]-'0';
  24. 	top=top-1;
  25. 	return  num;
  26. }
  27.  
  28. void push(char strdata){
  29. 	data[++top]=strdata;
  30. }
  31.  
  32.  
  33.  
  34. int main(int argc, char *argv[]) {
  35. 	char str[25];
  36. 	int f=0,i=0;//f為旗標 
  37. 	char token='\0';
  38. 	int ans,length;
  39.  
  40.     puts("請輸入字串：");
  41.     gets(str);
  42.     length = strlen(str);
  43.     puts("輸入的字串為：");
  44.     printf("字串=%s ; 長度=%d \n",str,length);
  45.     printf("-----------------------\n");
  46.  
  47.     while(f==0){
  48.     	if(str[i]>='0'&& str[i]<='9'){//若是數字丟入運算元陣列 
  49.     		push(str[i]);
  50.     		i++;//i計算算式字串位置 
  51.     		continue;
  52. 		}else if(str[i]=='*' || str[i]=='/' || str[i]=='+' || str[i]=='-'){
  53. 			if(str[i]=='+' || str[i]=='-'){//若是運算子，先判斷optop運算子堆疊是否有內容 
  54. 				if(optop>=0){//有內容才判斷優先權 
  55. 					if(opstack[optop]=='*' || opstack[optop]=='/'){
  56. 						token=oppop();
  57. 						switch(token){
  58. 							case '*':
  59. 								ans=pop()*pop();
  60. 								push(ans);
  61. 								break;
  62. 							case '/':
  63. 								ans=pop()/pop();
  64. 								push(ans);
  65. 								break;
  66. 						}
  67. 						oppush(str[i]);	
  68. 						i++;	
  69. 					}else{//若堆疊內運算子是+-則直接放入 
  70. 						oppush(str[i]);
  71. 						i++;
  72. 					}	
  73. 				}else{//運算子堆疊無內容直接放入 
  74. 						oppush(str[i]);
  75. 						i++;
  76. 					}	
  77. 			}else{//判斷*/優先權 
  78. 				if(optop>=0){//同上方 
  79. 					if(opstack[optop]=='*' || opstack[optop]=='/'){
  80. 						token=oppop();
  81. 						switch(token){
  82. 							case '*':
  83. 								ans=pop()*pop();
  84. 								push(ans);
  85. 								break;
  86. 							case '/':
  87. 								ans=pop()/pop();
  88. 								push(ans);
  89. 								break;
  90. 						}
  91. 						oppush(str[i]);	
  92. 						i++;	
  93. 					}else{
  94. 						oppush(str[i]);
  95. 						i++;
  96. 					}
  97. 				}else{
  98. 						oppush(str[i]);
  99. 						i++;
  100. 					}	
  101. 			}	
  102. 		}else{
  103. 			f=1;
  104. 		}
  105. 	}
  106.  
  107. 	while(optop>=0){//最後拿出剩下的運算元堆疊與運算子堆疊倆倆運算 
  108. 		token=oppop();
  109. 		switch(token){
  110. 			case '+':
  111. 				ans=pop()+pop();
  112. 				push(ans);
  113. 				break;
  114. 			case '-':
  115. 				ans=pop()+pop();
  116. 				push(ans);
  117. 				break;
  118. 			case '*':
  119. 				ans=pop()*pop();
  120. 				push(ans);
  121. 				break;
  122. 			case '/':
  123. 				ans=pop()/pop();
  124. 				push(ans);
  125. 				break;
  126. 			default:
  127. 				ans=pop();
  128. 				break;
  129. 		}
  130. 		optop--;
  131. 	}
  132.  
  133. 	printf("答案是%d \n",ans);
  134.  
  135.  
  136.  
  137. 	return 0;
  138. }
Success #stdin #stdout 0s 2164KB
comments (?)
stdin
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Standard input is empty
stdout
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請輸入字串：
輸入的字串為：
字串=�rﯿ��rﯿ/ ; 長度=14 
-----------------------
答案是-1
https://ideone.com/rjbx8a
language:
C (gcc 8.3)
created:
9 years ago
visibility:
public

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