如何插入，将新元素添加到char数组中

Question

我推送新元素到char数组前面有问题。我在我的程序中有一个函数，它有两个参数：一个指向char数组的指针和一个指向int的指针。该函数用一些数据填充数组，并返回此数组中元素的数量。

int function1(char* buffer , int* outSize); 

buffer[1000] 
int size; 
result = function1(buffer , &size); 

// after this a get some data in buffer and count how many data is in buffer. let's say 456 elements 
// now a try push to front two new bytes as size this array 

finallBuffer *char = new char[size+2]; 
memcpy(finallyBuffer, (char*)&size, 2); // here a get size in two bytes and put to finnalBuffer 

// next I try just copy to finnalyBuffer received data from Buffer 
strcat(finallyBuffer , buffer); 
doSomething(finallyBuffer); 
delete []finallyBuffer; 

在此之后在finnalyBuffer我只保存大小两个字节。为什么我看不到来自缓冲区的数据？

在最后我要实现新表在前面的两个新的字节。这两个字节是大小旧的表。比方说，接收到的数据HEVE只有5个字符

char table[5] = {'a','b','b','c','t'};

所以尺寸为5.5两个字节是char size = {'5','0'};

结果应该是的样子。

char table[5] = {'5','0','a','b','b','c','t'};

Answer 1

我不知道确切地知道你想要做什么，但我试图理解它。在代码见下面的意见，我希望它会帮助你。

int function1(char* buffer , int* outSize); 

buffer[1000] 
int size; 
result = function1(buffer , &size); 

// after this a get some data in buffer and count how many data is in buffer. let's say 456 elements 
// now a try push to front two new bytes as size this array 

char *finallBuffer = new char[size*sizeof(char)+2]; 
short ssize = (short)size; 
memcpy(finallyBuffer, &ssize, 2); // here a get size in two bytes and put to finnalBuffer 

// next I try just copy to finnalyBuffer received data from Buffer 

/* NOTE: strcat is for null terminated string. If i understand what you want 
    to do you want store int in 4 bytes ? 
    I assume size is the actual size of the buffer 
*/ 
//strcat(finallyBuffer , buffer); 
memcpy(&finallyBuffer[2], buffer, sizeof(char)*size); 
doSomething(finallyBuffer); 
delete []finallyBuffer; 

Answer 2

那些线接错：

finallBuffer *char = new char[size+2]; 
memcpy(finallyBuffer, (char*)&size, 2); 

他们应该是：

char * finalBuffer = new char[size+2]; 
memcopy(finallyBuffer, buffer, size); 

但我真的理解你正在尝试做一些事情。

编辑：

现在我看到你在做什么，奥利弗给了你答案。如果你不想（或不能）为你的缓冲区使用std :: vector，那么无论如何你都可以节省重新分配数组的代价。

• 刚开始为缓冲区2分配更多字节：char buffer[1002];。
• 然后要存储值，您可以将指针传递给第三个元素：char* ptr = buffer + 2;。
• 当你知道这些数据的实际尺寸，只是把字节buffer[0]和buffer[1]。

一个结构应建议。