Question

gcc 4.4.4 c89

我的channel.h文件中有以下代码

typedef struct channel_tag channel_t;

channel_t* open_channel(size_t channel_id);
void close_channel(channel_t *channel);

在我的channel.c文件中

#include "channel.h"

struct channel_tag {
    size_t channel_id;
};

channel_t* open_channel(size_t channel_id)
{
    channel_t *channel = malloc(sizeof *channel);

    if(channel == NULL) {
        fprintf(stderr, "Cannot allocate memory\n");
        return NULL;
    }

    channel->channel_id = channel_id;
    printf("Channel [ %zu ] has been created\n", channel->channel_id);

    return channel;
}

void close_channel(channel_t *channel)
{
    printf("Channel [ %zu ] resources has been released\n", channel->channel_id);
    free(channel);
}

问题在于我的main.c文件。这里我有一个for循环，它创建5个通道对象并为它们分配内存。但是，如果我想在我的程序中稍后释放它们，我不确定如何能够引用它们。这只是我正在测试的5。但后来可能会达到300个。

int main(void)
{
    size_t i = 0;

    channel_t *channel = NULL;

    for(i = 0; i < 4; i++) {
        channel = open_channel(i);

        if(channel == NULL) {
            fprintf(stderr, "Cannot create channel [ %zu ]\n", i);
        }
    }

    /* Do some stuff with the channels and now free them before the program exists. 
       However, I need to loop and pass all of them, not just one */
    for(i = 0; i < 4; i++) {
       close_channel(channel);
    }
    return 0;
}

非常感谢任何建议，

Answer 1

好吧，你在主要部分反复重写同一个频道。如果你想要4个通道，你显然需要4个变量来存储它们，或者一个包含4个项目的数组。

channel_t *channel[4]; for (...) channel[i] = open_channel(i);

哦，for (int i = 0; i < 4; i++)将产生4个循环，而不是5个。

Answer 2

在创建通道时将通道存储在数组中。确保您可以在程序结束时判断malloc是否有效（因此此代码中为memset）。

channel_t **channel = malloc(5 * sizeof(channel_t*));
memset(channel, 0, 5 * sizeof(channel_t*));

for(i = 0; i < 5; i++) {
    channel[i] = open_channel(i);

    if(channel[i] == NULL) {
        fprintf(stderr, "Cannot create channel [ %zu ]\n", i);
    }
}

/* Do some stuff with the channels and now free them before the program exists. 
   However, I need to loop and pass all of them, not just one */
for(i = 0; i < 5; i++) {
   if (channel[i] != NULL)  /* Handle case where some of the opens failed */
   {
     close_channel(channel[i]);
   }
}

free(channel);

Answer 3

int main( void )
{
    int i = 0;
    channel_t** channels = malloc( 5 * sizeof( channel_t* ));

    if ( channels == NULL ) exit( 1 ); // error, etc.

    for ( i = 0; i < 5; i++ ) {
        channels[i] = open_channel( i );

        if ( channels[i] == NULL ) {
            fprintf( stderr, "Cannot create channel [ %d ]\n", i );
            break;
        }
    }

    // ... work, but only if i == 5

    for ( i--; i >= 0; i-- ) {
       close_channel( channels[i] );
    }

    free( channels );
    return 0;
}

分配和释放结构的对象

3 个答案: