发布时间 : 星期一 文章长沙理工大学数据结构栈的实现及应用算术表达式求值实验报告更新完毕开始阅读4a5db14d30b765ce0508763231126edb6f1a76d1
//判断TypeStack是否为满 bool IsFull_Type(Stack S) { }
void Push_Char(Stack *S, char ch) { }
void Push_Type(Stack *S, Type a) { }
char Pop_Char(Stack *S)
//TypeStack不为满则入栈，否则输出提示 if(!IsFull_Type(*S))
S->TypeStack[S->TypeTop++] = a; //charStack不为满则入栈，否则输出提示 if(!IsFull_Char(*S))
S->charStack[S->charTop++] = ch; return S.TypeTop == MAX;
else
cout << \
else
cout << \
{ }
Type Pop_Type(Stack *S) { }
char Top_Char(Stack S) {
if(!IsEmpty_Char(S)) if(!IsEmpty_Type(*S)) { } else
cout << \S->TypeTop--;
return S->TypeStack[S->TypeTop]; if(!IsEmpty_Char(*S)) { } else
cout << \S->charTop--;
return S->charStack[S->charTop];
return -1;
return -1;
}
return S.charStack[--S.charTop];
else
cout << \
return -1;
Type Top_Type(Stack S) { }
Type Calculate(Type left, Type right, char op) {
Type value = 0; switch(op) {
case '+': value = left + right; break; case '-': value = left - right; break; case '*': value = left * right; break; case '/': if(right != 0)
value = left / right;
if(!IsEmpty_Type(S))
return S.TypeStack[--S.TypeTop];
else
cout << \
return -1;
else
}
cout << \被除数不能为零!\
break;
case '%': if(right != 0)
value = (int)left % (int)right;
else
cout << \被余数不能为零!\
break;
case '^': value = pow(left,right); /*value = 1; }
return value;
if(right >= 0)
while(right--)
value *= left;
else { }*/
right = -right; while(right--)
value /= left;
void Computer(char *mid_equotion, Type len) {
Type right, left , result;