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我正在尝试C 语言中的协议缓冲区序列化和反序列化。我创建了如下所示的 learn.proto 消息，并尝试将消息打包到 *uint8_t 中并解压缩。

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，无论我将什么打包到原始消息中，都会获得始终为零的枚举值。但是反序列化后文本字段值仍然正确。例如，我正在打包枚举LEARN__LEARNMESSAGE__LEARN_TYPE__THREE ，但解包后，它变为LEARN__LEARNMESSAGE__LEARN_TYPE__ZERO

我编译它gcc -o program mycode.c learn.pb-c.c -lprotobuf-c

有人可以帮我找出我正在做的错误：

#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)


}
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1 回答 1

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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";
于 2017-03-23T09:18:40.720 回答