求加减乘除c语言代码 用c语言编一个字符型计算器代码（只要求有加减乘除，并且能分清...
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#include
int main()
{
int a,b;
scanf("%d %d",&a,&b);
printf("%d\t",a + b);
printf("%d\t",a - b);
printf("%d\t",a * b);
printf("%d\t",a / b);
return 0;
}
\u8fd0\u884c\u622a\u56fe
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#include
struct s_node
{
int data;
struct s_node *next;
};
typedef struct s_node s_list;
typedef s_list *link;
link operator=NULL;
link operand=NULL;
link push(link stack,int value)
{
link newnode;
newnode=(link) malloc(sizeof(s_list));
if(!newnode)
{
printf("\nMemory allocation failure!!!");
return NULL;
}
newnode->data=value;
newnode->next=stack;
stack=newnode;
return stack;
}
link pop(link stack,int *value)
{
link top;
if(stack !=NULL)
{
top=stack;
stack=stack->next;
*value=top->data;
free(top);
return stack;
}
else
*value=-1;
}
int empty(link stack)
{
if(stack==NULL)
return 1;
else
return 0;
}
int is_operator(char operator)
{
switch (operator)
{
case '+': case '-': case '*': case '/': return 1;
default:return 0;
}
}
int priority(char operator)
{
switch(operator)
{
case '+': case '-' : return 1;
case '*': case '/' : return 2;
default: return 0;
}
}
int two_result(int operator,int operand1,int operand2)
{
switch(operator)
{
case '+':return(operand2+operand1);
case '-':return(operand2-operand1);
case '*':return(operand2*operand1);
case '/':return(operand2/operand1);
}
}
void main()
{
char expression[50];
int position=0;
int op=0;
int operand1=0;
int operand2=0;
int evaluate=0;
printf("\nPlease input:\n");
gets(expression);
while(expression[position]!='\0'&&expression[position]!='\n')
{
if(is_operator(expression[position]))
{
if(!empty(operator))
while(priority(expression[position])data)&&
!empty(operator))
{
operand=pop(operand,&operand1);
operand=pop(operand,&operand2);
operator=pop(operator,&op);
operand=push(operand,two_result(op,operand1,operand2));
}
operator=push(operator,expression[position]);
}
else
operand=push(operand,expression[position]-48);
position++;
}
while(!empty(operator))
{
operator=pop(operator,&op);
operand=pop(operand,&operand1);
operand=pop(operand,&operand2);
operand=push(operand,two_result(op,operand1,operand2));
}
operand=pop(operand,&evaluate);
printf("The expression [%s] result is '%d' ",expression,evaluate);
getch();
}
直接 变量名+运算符+变量名就可以了啊。
注意除法 /是整除符号，如果符号两边都是整数（int型），那么结果值保留整数，小数将被忽略。 如果要精确到小数，'/'两边至少有一个是浮点数（float或double）。
注意混合运算结果是按照表达式中数据表达范围最大的数据类型，例如：整数和浮点数混合运算，结果返回是浮点数。
double a,b;
char c;
double result;
scanf ("%lf %c %lf",&a,&c,&b);
switch (c)
{
case '+': result=a+b;break;
case '-': result=a-b;break;
case '*': result=a*b;break;
case '/': result=a/b;break;
}
int a,b,c
c = a+b
有难度?
在百度文库搜“C语言简易四则运算计算器代码”。
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