c - 如何解决c警告:pointer from integer without a cast?

Question

警告:传递‘memcpy’的参数 2 使指针来自整数而不进行强制转换[默认启用]

uint8 InterruptLatency;
char buf[256];
int kernelinterrupt time()
{
    fscanf(fp, "%"SCNu8"\n", &InterruptLatency);  // I am reading the data from kernel which is not shown here
  memcpy(buf, InterruptLatency, sizeof(InterruptLatency));   // warning here as above

    // after storing it in buffer I am sending the data from but to another layer
}

Answer 1

memcpy() 函数需要两个指针，但 InterruptLatency 是一个 8 位整数。

解决方法是取变量的地址:

memcpy(buf, &InterruptLatency, sizeof InterruptLatency);
            ^
            |
        address-of
        operator

请注意，在获取实际对象的大小时，sizeof 不需要括号。这是因为 sizeof 不是函数。

另请注意，像这样使用memcpy() 将单个字节复制到字节数组中永远不会发生在“真正的”C 代码中。我会这样做:

buf[0] = InterruptLatency;