Question

我正在尝试协议缓冲序列化＆amp; C language中的反序列化。我创建了一个learn.proto消息，如下所示，只是尝试将消息打包到* uint8_t并解压缩。

syntax = "proto3";
package learn;

message learnmessage {

  enum LearnType {
        ZERO = 0;
        ONE = 1;
        TWO = 2;
        THREE = 3;
        FOUR = 4;
        FIVE = 5;
        SIX = 6;
    }
    LearnType typ = 1;
    string text = 2;
}

解包后，无论我打包到原始消息中，都会获得枚举值，始终为零。但反序列化后文本字段值仍然正常。例如，我正在打包enum LEARN__LEARNMESSAGE__LEARN_TYPE__THREE 但在解压后，它会变为LEARN__LEARNMESSAGE__LEARN_TYPE__ZERO

我按gcc -o program mycode.c learn.pb-c.c -lprotobuf-c

编译

有人可以帮助我弄清楚我正在做的错误：

#include <stdio.h>
#include <stdlib.h>
#include "learn.pb-c.h"


int main()
{
    Learn__Learnmessage msg = LEARN__LEARNMESSAGE__INIT, *msgCopy; 
    size_t len;
    uint8_t *msgStr;

    msg.typ = LEARN__LEARNMESSAGE__LEARN_TYPE__THREE; //corresponds to enum value 3
    msg.text = "again";

    len = learn__learnmessage__get_packed_size(&msg); //get the length of the uint8_t needed to pack the message
    msgStr = malloc(len * sizeof(uint8_t));

    learn__learnmessage__pack(&msg, msgStr); //packs the message into msgStr

    msgCopy = learn__learnmessage__unpack(NULL, len, msgStr); //unpack the message into msgCopy

    printf("Orginal: %d \n", msg.typ);  //prints 3 (expected)
    printf("Copy: %d \n", msgCopy->typ); //prints 0 (un-expected)


}

Answer 1

生成的C代码中有一个字段protobuf_c_boolean has_typ;。将其设置为1后，事情开始按预期工作。

所以代码实际上必须是：

msg.typ = LEARN__LEARNMESSAGE__LEARN_TYPE__THREE;
msg.has_typ = 1;
msg.text = "again";

协议消息的反序列化不正确

1 个答案: