C语言转移表之加减乘除无限进化版

主干程序

• 输入程序
• 解析程序

/*解析字符串 有空格把空格分开
比如输入：add 1 2
解析后：add
       1
       2
 */
void do_parse(char *buf)
{
    int state = 0;
    int i = 0;
    int argc = 0;
    char *argv[8] = {0};
    for (i = 0; buf[i]; i++)
    {
        if (state == 0 && !isspace(buf[i]))
        {
            state = 1;
            argv[argc++] = buf + i;
        }
        else if(isspace (buf[i]))
        {
            state = 0;
            buf[i] = 0;
        }
    }
    argv[argc] = NULL;
    /*核心算法*/
    do_math(argc, argv);
}

int main()
{
    char buf[1000] = {0};
    while (1)
    {
        printf(">");
        scanf("%[^\n]%*c", buf);//输入
        if (strcmp(buf, "exit") == 0)//如果输入了"exit"就结束
            break;
        else
            do_parse(&buf);//解析程序
    }
}

初级版本

void do_math(char * argc, char *argv[])
{
    if (strcmp(argv[0], "add") == 0)
    {
        int a = atoi(argv[1]);//字符串转a整形
        int b = atoi(argv[2]);//字符串转b整形
        printf("%s + %s = %d\n", argv[1], argv[2], add(a, b));//打印加法结果a+b
    }
    else if (strcmp(argv[0], "sub") == 0)
    {
        int a = atoi(argv[1]);
        int b = atoi(argv[2]);
        printf("%s - %s = %d\n", argv[1], argv[2], sub(a, b));
    }
}

进阶版本

int add(int a, int b)
{
    return a + b;
}
int sub(int a, int b)
{
    return a - b;
}

int(*pf)(int a, int b) = NULL;
void do_math(char * argc, char *argv[])
{
    if (strcmp(argv[0], "add") == 0)
    {
        pf = add;
    }
    if (strcmp(argv[0], "sub") == 0)
    {
        pf = sub;
    }
    int a = atoi(argv[1]);
    int b = atoi(argv[2]);
    printf("%s + %s = %d\n", argv[1], argv[2], pf(a, b));
}

#include<stdio.h>
#include<stdlib.h>

float add(float a, float b)
{
    return a + b;
}
float sub(float a, float b)
{
    return a - b;
}
float mul(float a, float b)
{
    return a * b;
}
float div(float a, float b)
{
    return a / b;
}


int main(void)
{
    float(*operate[])(float, float) = { add,sub,mul,div };

    float number_a, number_b;
    float result;
    int oprate;
    printf("input the operator!1 +,2  -,3  *,4  /\n");
    scanf("%d", &oprate);
    printf("Input the number_a:");
    scanf("%f", &number_a);
    printf("Input the number_b:");
    scanf("%f", &number_b);

    printf(
        "the result is %.3f\n",
        result = operate[oprate - 1](number_a, number_b)
    );

    system("pause");
    return 0;
}

这两句关键程序

float(*operate[])(float, float) = { add,sub,mul,div };
result = operate[oprate - 1](number_a, number_b)

版本进化