将图像从C#客户端发送到C服务器
如果我发送纯文本,没有问题。一切都好。 但是,如果我尝试从C#client发送一个映像,服务器会收到正确的字节数,但是当我将缓冲区保存到文件(以二进制模式-wb)时,它总是有4个字节 我使用函数将图像从C#客户端发送到C服务器,c#,c,sockets,C#,C,Sockets,如果我发送纯文本,没有问题。一切都好。 但是,如果我尝试从C#client发送一个映像,服务器会收到正确的字节数,但是当我将缓冲区保存到文件(以二进制模式-wb)时,它总是有4个字节 我使用函数File.ReadAllBytes()通过C#客户端发送它 我的保存代码看起来像 FILE * pFile; char *buf = ReceiveMessage(s); pFile = fopen (fileName , "wb"); fwrite(buf, sizeof(buf
File.ReadAllBytes()
通过C#客户端发送它
我的保存代码看起来像
FILE * pFile;
char *buf = ReceiveMessage(s);
pFile = fopen (fileName , "wb");
fwrite(buf, sizeof(buf[0]), sizeof(buf)/sizeof(buf[0]), pFile);
fclose (pFile);
free(buf);
static unsigned char *ReceiveMessage(int s)
{
int prefix;
recv(s, &prefix, 4, 0);
int len = prefix;
char *buffer= (char*)malloc(len + 1);
int received = 0, totalReceived = 0;
buffer[len] = '\0';
while (totalReceived < len)
{
if (len - totalReceived > BUFFER_SIZE)
{
received = recv(s, buffer + totalReceived, BUFFER_SIZE, 0);
}
else
{
received = recv(s, buffer + totalReceived, len - totalReceived, 0);
}
totalReceived += received;
}
return buffer;
}
我的接收函数看起来像
FILE * pFile;
char *buf = ReceiveMessage(s);
pFile = fopen (fileName , "wb");
fwrite(buf, sizeof(buf[0]), sizeof(buf)/sizeof(buf[0]), pFile);
fclose (pFile);
free(buf);
static unsigned char *ReceiveMessage(int s)
{
int prefix;
recv(s, &prefix, 4, 0);
int len = prefix;
char *buffer= (char*)malloc(len + 1);
int received = 0, totalReceived = 0;
buffer[len] = '\0';
while (totalReceived < len)
{
if (len - totalReceived > BUFFER_SIZE)
{
received = recv(s, buffer + totalReceived, BUFFER_SIZE, 0);
}
else
{
received = recv(s, buffer + totalReceived, len - totalReceived, 0);
}
totalReceived += received;
}
return buffer;
}
static unsigned char*ReceiveMessage(int-s)
{
int前缀;
recv(s和前缀,4,0);
int len=前缀;
char*buffer=(char*)malloc(len+1);
接收的int=0,接收的totalReceived=0;
缓冲区[len]='\0';
while(总接收缓冲区大小)
{
接收=recv(s,缓冲区+接收总数,缓冲区大小,0);
}
其他的
{
received=recv(s,缓冲区+totalReceived,len-totalReceived,0);
}
totalReceived+=已接收;
}
返回缓冲区;
}
您犯了一个初学者错误,使用了sizeof(buf)
。它不返回缓冲区中的字节数,而是返回指针的大小(根据您运行的是32位还是64位平台,指针的大小是四还是八)
您需要更改
ReceiveMessage
函数以同时“返回”所接收数据的大小。您的C代码需要从ReceiveMessage()函数返回len
char *buf = ReceiveMessage(s); // buf is a char*
... sizeof(buff) // sizeof(char*) is 4 or 8
所以你需要像这样的东西
static unsigned char *ReceiveMessage(int s, int* lenOut)
{
...
*lenOut = totalReceived ;
}
您不能通过
sizeof
获得数组的大小。更改为,即:
int len = 0;
char *buf;
buf = ReceiveMessage(s, &len);
/* then use len to calculate write length */
static unsigned char *ReceiveMessage(int s, int *len)
/* or return len and pass ptr to buf */
{
...
}
为什么我不能使用
strlen(buf)
而不是?因为strlen()
是用于字符串的,并且只计算到第一个'\0'
为什么我不能使用strlen(buf)
而不是?@daroPL可以,但前提是接收的数据实际上是正确的字符串。如果是图像,则它包含二进制数据,其中可以包含一个值为0
的字节,该字节用作字符串终止符。