c - How can I resolve this warning?

Question

The code compiles and runs but I want to handle the warning.

#include <stdio.h>
#include "includes.h"
#include <string.h>

#define DEBUG 1

/* Definition of Task Stacks */
/* Stack grows from HIGH to LOW memory */
#define   TASK_STACKSIZE       2048
OS_STK    task1_stk[TASK_STACKSIZE];
OS_STK    task2_stk[TASK_STACKSIZE];
OS_STK    stat_stk[TASK_STACKSIZE];

/* Definition of Task Priorities */
#define TASK1_PRIORITY      6  // highest priority
#define TASK2_PRIORITY      7
#define TASK_STAT_PRIORITY 12  // lowest priority 

void printStackSize(INT8U prio)
{
    INT8U err;
    OS_STK_DATA stk_data;

    err = OSTaskStkChk(prio, &stk_data);
    if (err == OS_NO_ERR) 
    {
        if (DEBUG == 1)
           printf("Task Priority %d - Used: %d; Free: %d\n", 
                   prio, stk_data.OSFree, stk_data.OSUsed);
    }
    else
    {
        if (DEBUG == 1)
           printf("Stack Check Error!\n");    
    }
}

Answer 1

不知道INT32U它的定义很难说，但你可能想要：

printf("Task Priority %d - Used: %u; Free: %u\n", 
        prio, stk_data.OSFree, stk_data.OSUsed);

或者：

printf("Task Priority %d - Used: %lu; Free: %lu\n", 
        prio, stk_data.OSFree, stk_data.OSUsed);

Answer 2

我相信编译器试图告诉你你的 printf 格式字符串需要一个有符号整数（%d），但你给它的是一个无符号整数。如果是这样，您应该能够通过使用 %u （而不是 %d）或将您的值转换为 (int) 来消除警告（如果值大于有符号的 int 可能会成为问题）按住您的系统...这将导致显示负值）。

Answer 3

你可以用inttypes.h. 此标头定义固定整数的所有打印和扫描说明符。

所以这

printf("Task Priority %d - Used: %u; Free: %u\n", 
    prio, stk_data.OSFree, stk_data.OSUsed);

变成

printf("Task Priority %d - Used: %" PRIu32 "; Free: %" PRIu32 "\n", 
    prio, stk_data.OSFree, stk_data.OSUsed);