C 打印 void * 缓冲区

Question

好吧，这个

void printPacketBuffer(void *buffer, unsigned int length)
{
    unsigned int i=0;
    char *c;
    unsigned int limit = ( (length != 0)?length:PACKET_DATA );
    for (i=0; i<limit; i++) {
        c = (char *)(buffer+i);
        if ( *c != '\0' ) {
            printf("%c", *c);
        }
    }
}

正在工作，但以下函数缺少最后 14 个字节，无法理解为什么？!

void printPacket (void * buffer) {
    unsigned int size = getPacket_size(buffer);
    printf("\n***********\nNew Packet holding %d bytes of data\n***********\n", size);
    char *c;
    int counter=0;
    int i;
    for (i=2; i<size+2; i++) {
        c = (char *)(buffer+i);
        if ( *c != '\0' ) {
            printf("%c", *c);
            counter++;
        }
    }
    printf("\nactual printing %d, i=%d\n", counter, i);
}

这是通话

printPacketBuffer(pbuffer+2, getPacket_size(pbuffer));
printPacket(pbuffer);

Answer 1

我认为 pbuffer+2 是两个字节的下一个地址，但我错了，这就是发生的事情如果 bpuffer = 0x7fff2c03d8a0 则 bpuffer + 2 = 0x7fff2c03d8b0现在移动到我使用的下一个 2 字节位置

void *temp = (char *)pbuffer+shift;

无论如何，这个故事的主旨是使用字节转换为无符号字符

void printPacket (void * buffer) {
    unsigned int size = getPacket_size(buffer);
    printf("\n***********\nNew Packet holding %d bytes of data %p\n***********\n", size, buffer);
    char *c;
    int counter=0;
    int i;
    for (i=2; i<size+2; i++) {
        c = (char *)buffer+i;
        if ( *c != '\0' ) {
            printf("%c", *c);
            counter++;
        }
        else {
            printf("NULL ");
        }
    }
}