你好，我正在学习指针，所以我创建了一个计算器。我设法从函数返回值和指针，但通过全局声明它们。如何在本地声明它们？
#include <stdio.h>
#include <stdlib.h>
所有函数的声明
int Addition();
int Subtraction();
int Devision();
int Multiplication();
全局变量声明，我想在本地声明它们
int p;
int n;
int *r=&n;
int *b=&p;
主功能开始
int main()
{
int g,s;
while (1)
{
printf("Please choose the Arithmetic operation: n");
printf("Addition-> 1 nSubtraction-> 2 nDevision-> 3 nMultiplication-> 4 nExit-> 0n ");
scanf("%d",&g);
用户通过输入一个数字(1,2,3,4或0表示退出(来选择算术运算(函数(
if (g==1)
{
s=Addition(r,b);
printf("The addition result is %d+%d=%d",*r, *b, s);
}
else if (g==2)
{
s=Subtraction(r,b);
printf("The Subtraction result is %d-%d=%d",*r, *b, s);
}
else if (g==3)
{
s=Devision(r,b);
printf("The Devision result is %d/%d=%d",*r, *b, s);
}
else if (g==4)
{
Multiplication(r,b);
printf("The Multiplication result is %d/%d=%d",*r, *b, s);
}
else
{
break;
}
return 0;
}
}
主功能结束。以下是所有其他功能
Addition()
{
int x;
printf("Ennter first nr: ");
scanf("%d",&n);
printf("Ennter second nr: ");
scanf("%d",&p);
x=n+p;
return x;
}
Subtraction()
{
int x;
printf("Ennter first nr: ");
scanf("%d",&n);
printf("Ennter second nr: ");
scanf("%d",&p);
x=n-p;
return x;
}
Devision()
{
int x;
printf("Ennter first nr: ");
scanf("%d",&n);
printf("Ennter second nr: ");
scanf("%d",&p);
x=n / p;
return x;
}
Multiplication()
{
int x;
printf("Ennter first nr: ");
scanf("%d",&n);
printf("Ennter second nr: ");
scanf("%d",&p);
x=n * p;
return x;
}
1( 如果需要，请刷新有关指针的知识并通过pointers传递参数。
间接或取消引用运算符*提供指针指向的对象的内容。
元或一元运算符&给出变量的地址。
2(避免全局变量在main后在本地声明它们。
3( 使用switch而不是if-else链
#include <stdio.h>
#include <stdlib.h>
int Addition(int* n, int* p)
{
int x;
printf("Ennter first nr: ");
scanf("%d",n);
printf("Ennter second nr: ");
scanf("%d",p);
x = *n + *p;
return x;
}
int Subtraction(int* n, int* p)
{
int x;
printf("Ennter first nr: ");
scanf("%d",n);
printf("Ennter second nr: ");
scanf("%d",p);
x = *n - *p;
return x;
}
int main(void)
{
int n1 = 0;
int n2 = 0;
int *r = &n1;
int *b = &n2;
int g,s;
while (1)
{
printf("Please choose the Arithmetic operation: n");
printf("Addition-> 1 nSubtraction-> 2 nDevision-> 3 nMultiplication-> 4 nExit-> 0n ");
scanf("%d",&g);
switch(g)
{
case 0:
printf("ENDn");
return 0;
break;
case 1:
s = Addition(r, b);
printf("The addition result is: %d+%d=%dnn",n1, n2, s);
break;
case 2:
s = Subtraction(r ,b);
printf("The Subtraction result is: %d-%d=%dnn",*r, *b, s);
break;
}
}
return 0;
}
测试：
Please choose the Arithmetic operation:
Addition-> 1
Subtraction-> 2
Devision-> 3
Multiplication-> 4
Exit-> 0
2
Ennter first nr: 5
Ennter second nr: 3
The Subtraction result is: 5-3=2
Please choose the Arithmetic operation:
Addition-> 1
Subtraction-> 2
Devision-> 3
Multiplication-> 4
Exit-> 0
0
END
我在全球范围内使用它们，因为我更容易调用和阅读它们。这就是为什么我想学习如何在本地声明它们。
很简单，只需在本地声明它们，如下所示：(无需用户输入代码即可专注于您的主要问题(
int main(void)
{ // locally declared
double a = 4.5;
double b = 10.8;
double result_1, result_2, result_3, result_4;
result_1 = add(a, b);
result_2 = sub(a, b);
result_3 = mul(a, b);
result_4 = div(a, b);
return 0;
}
// example function (the others will be of similar form,
// a single return line with the appropriate math operator.
double add(double a, double b)//no pointers necessary
{
return a + b;
}
如果你真的想使用指针，那么你可以实现函数通过参数返回结果：
void add(double a, double b, double *result)
{
*result = a + b;
}
用法：(例如从主调用(
add(a, b, &result_1);// passing the address of result_1
// to allow the value at that address
// to be modified.
好吧，不确定这是否是你想要的，但让我建议你这段代码：
#include <stdio.h>
#include <stdlib.h>
int Addition(int* n, int* p)
{
int x;
printf("Ennter first nr: ");
scanf("%d",n);
printf("Ennter second nr: ");
scanf("%d",p);
x=*n+*p;
return x;
}
int main()
{
int n1=0;
int n2=0;
int *r=&n1;
int *b=&n2;
int g,s;
while (1)
{
printf("Please choose the Arithmetic operation: n");
printf("Addition-> 1 nSubtraction-> 2 nDevision-> 3 nMultiplication-> 4 nExit-> 0n ");
scanf("%d",&g);
if (g==1)
{
s=Addition(r,b);
printf("The addition result is %d+%d=%d",*r, *b, s);
}
return 0;
}
}
如您所见，我们使用指针来存储用户写入的数字的值，因此您可以在调用函数后保留它们。