Question

我有这个结构：

struct ChangeIntItem
{
    char *unit;
    const char **parser;
    int *changevalue;
    uint16_t *change_eeprom_value;
    int maximum;
    int minimum;
};

我想用这个struct-Variable初始化其他变量：

struct ChangeIntItem ChangeIntItemTypeBoolean = { .unit = "", .minimum = 0, .maximum = 1, .parser = {"off", "on"}};

工作正常，但我收到一些警告：

Severity    Code    Description Project File    Line
Warning     braces around scalar initializer    Handsteuerung   C:\Users\...    11

Severity    Code    Description Project File    Line
Warning     (near initialization for 'ChangeIntItemTypeBoolean.parser') Handsteuerung   C:\Users\...    11

Severity    Code    Description Project File    Line
Warning     initialization from incompatible pointer type   Handsteuerung   C:\Users\...    11

Severity    Code    Description Project File    Line
Warning     (near initialization for 'ChangeIntItemTypeBoolean.parser') Handsteuerung   C:\Users\...    11

Severity    Code    Description Project File    Line
Warning     excess elements in scalar initializer   Handsteuerung   C:\Users\...    11

Severity    Code    Description Project File    Line
Warning     (near initialization for 'ChangeIntItemTypeBoolean.parser') Handsteuerung   C:\Users\...    11

在另一个案例中，我写了一个函数，它将struct的变量设置为default-values，但我更喜欢这个方法，因为它更短。

由＆＃39; .parser = {＆＃34; off＆＃34;，＆＃34; on＆＃34;}＆＃39;引起的所有错误但我没有弄错...

Answer 1

您可以使用compound literals，因此请将init更改为

struct ChangeIntItem ChangeIntItemTypeBoolean = { .unit = "", .minimum = 0, .maximum = 1, .parser = (const char *[]){"off", "on"}};

测试

#include <stdio.h>
#include <stdint.h>

struct ChangeIntItem
{
    char *unit;
    const char **parser;
    int *changevalue;
    uint16_t *change_eeprom_value;
    int maximum;
    int minimum;
};

int main()
{
    struct ChangeIntItem ChangeIntItemTypeBoolean = { .unit = "", .minimum = 0, .maximum = 1, .parser = (const char *[]){"off", "on"}};

    printf ("%s - %s\n", ChangeIntItemTypeBoolean.parser[0], ChangeIntItemTypeBoolean.parser[1]);
}

输出

off - on

Answer 2

成员parser是一个指针。

将其更改为指针数组，因此初始化可以保持不变：

 const char *parser[2];

或使用复合文字或其他变量：

.parser = ( const char*[]){"off", "on"}

const char* array[2] = {"off", "on"};
.parser = array

结构初始化警告

2 个答案: