static调试发现的问题

先给出代码吧：

#include "led.h"
#include "delay.h"
#include "sys.h"
#include "usart.h"

// FreeRTOS head file, add here.
#include "FreeRTOS.h"
#include "task.h"
#include "queue.h"
#include "list.h"
#include "portable.h"
#include "FreeRTOSConfig.h"

void task2(void *pvParameters);

xTaskHandle xTask2Handle;

static const char * Task1 = "task1 is running \r\n";
static const char * Task2 = "task2 is running \r\n";


void callFunction(void)
{
    static int temp = 0;

    temp ++;
    printf("temp: %d\r\n", temp);
}

void task1(void * pvParameters)
{
    char i = 0;
    char *TaskName = (char *)pvParameters;

    while(1)
    {
        printf(TaskName);

        for(i =0; i<3; i++)
          callFunction();

        //xTaskCreate( task2, "Task2", configMINIMAL_STACK_SIZE, (void *)Task2, 2, &xTask2Handle);
        vTaskDelay(1000/portTICK_RATE_MS);
    }
}

int main(void)
{
    // board initialize.
    LED_Init();
    uart_init(115200);

    //create task here.
    xTaskCreate( task1, "Task1", configMINIMAL_STACK_SIZE, (void *)Task1, 1, NULL);

    // start scheduler now
    vTaskStartScheduler();

    return 0;
}

在task1中三次调用了callFunction()函数，那么temp会怎么打印呢？
是不是没调用一次就创建一个变量temp呢？

运行结果是：

再看一个例子：如果有两个任务都在调用函数callFunction（）,则temp会怎么打印呢？
给出代码：

#include "led.h"
#include "delay.h"
#include "sys.h"
#include "usart.h"

// FreeRTOS head file, add here.
#include "FreeRTOS.h"
#include "task.h"
#include "queue.h"
#include "list.h"
#include "portable.h"
#include "FreeRTOSConfig.h"

void task2(void *pvParameters);

static const char * Task1 = "task1 is running \r\n";
static const char * Task2 = "task2 is running \r\n";


void callFunction(void)
{
    static int temp = 0;

    temp ++;
    printf("temp: %d\r\n", temp);
}

void task1(void * pvParameters)
{
    char i = 0;
    char *TaskName = (char *)pvParameters;

    while(1)
    {
        printf(TaskName);

        for(i =0; i<3; i++)
            callFunction();

        //xTaskCreate( task2, "Task2", configMINIMAL_STACK_SIZE, (void *)Task2, 2, &xTask2Handle);
        vTaskDelay(1000/portTICK_RATE_MS);
    }
}





void task2(void *pvParameters)
{
    char i = 0;
    char *TaskName = (char *)pvParameters;

    while(1)
    {
        printf(TaskName);

        for(i =0; i<3; i++)
            callFunction();

        vTaskDelay(1000/portTICK_RATE_MS);
    }
}


int main(void)
{
    // board initialize.
    LED_Init();
    uart_init(115200);

    //create task here.
    xTaskCreate( task1, "Task1", configMINIMAL_STACK_SIZE, (void *)Task1, 1, NULL);
    xTaskCreate( task2, "Task2", configMINIMAL_STACK_SIZE, (void *)Task2, 1, NULL);

    // start scheduler now
    vTaskStartScheduler();

    return 0;
}

打印结果又是啥样呢？

看到这个结果，是不是就明白什么了吗？

如果用static定义了局部静态变量，则调用的函数，都是在共享这个变量。
如果没有用这个static修饰的话，则有啥区别呢？