xilinx zynq7020通过ps uart实现简单计算器

本设计仅用到了zynq的ps端。利用串口完成计算数据的输入输出实现简易计算器。

代码语言:javascript
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/*

  • helloworld.c: simple test application
  • This application configures UART 16550 to baud rate 9600.
  • PS7 UART (Zynq) is not initialized by this application, since
  • bootrom/bsp configures it to baud rate 115200

  • | UART TYPE BAUD RATE |

  • uartns550 9600
  • uartlite Configurable only in HW design
  • ps7_uart 115200 (configured by bootrom/bsp)
    */

#include <stdio.h>
#include "platform.h"
#include "xil_printf.h"
#include "math.h"

double calculator();

int main()
{
init_platform();

print(&#34;Hello World\n\r&#34;);
calculator();
cleanup_platform();
return 0;

}

double calculator()
{
// 分别存放第一个操作数和第二个操作数以及结果的变量
double x1,x2,result;

// 存放运算符的变量
char m;

while(1)
{
    printf(&#34;请输入第一个数:\n&#34;);
    // 下面这得注意,接收double型的数据得用lf%,接收float用f%
    scanf(&#34;%lf&#34;,&amp;x1);

    printf(&#34;请输入运算操作(+ - * /):\n&#34;);
    m = getchar();
    printf(&#34;\n&#34;);

    printf(&#34;请输入第二个数:\n&#34;);
    scanf(&#34;%lf&#34;,&amp;x2);

    switch(m)
    {
        case &#39;+&#39;:
            printf(&#34;加法\n&#34;);
            result = x1 + x2;
            printf(&#34;%lf + %lf = %lf\n&#34;,x1,x2,result);
            break;

        case &#39;-&#39;:
            printf(&#34;减法\n&#34;);
            result = x1 - x2;
            printf(&#34;%lf - %lf = %lf\n&#34;,x1,x2,result);
            break;

        case &#39;*&#39;:
            printf(&#34;乘法\n&#34;);
            result = x1 * x2;
            printf(&#34;%lf * %lf = %lf\n&#34;,x1,x2,result);
            break;

        case &#39;/&#39;:
            printf(&#34;除法\n&#34;);
            if(x2 == 0)
            {
                printf(&#34;除数不能为0.\n&#34;);
            }
            else
            {
                result = x1 / x2;
                printf(&#34;%lf / %lf = %lf\n&#34;,x1,x2,result);
            }
            break;

        default:
            break;
    }
}

return 0.0;

}

调试: