该警告通常在不同的编译器上显示为错误。 C++ 中 Warning: Cast From Pointer to Integer of Different Size 编译器发出警告和错误来通知程序员代码有问题。 错误禁止代码执行,而警告允许代码执行,但有时这种执行可能会导致一些内存处理问题。 因此,建议在执行代码之前清除所有警告和错误。 当您尝试将整数的值
警告信息 "cast from pointer to integer of different size [-Wpointer-to-int-cast]" 指的是在代码中进行了从指针类型到整数类型的转换,但这个转换在大小上是不匹配的。在不同的平台上,指针的大小(通常是32位或64位)和整数类型(如 int,long 等)的大小可能不同。因此,直接进行这样的转换可能会导致数据丢失...
修改成：p=(void *)((unsigned int)a + (unsigned int)b)；错误原因：首先一点，两个地址相加没有意义，地址加一个偏移量才有意义，你一行实质是两个地址相加；如果你想练习使用指针访问变量的值，从而得到1000+11的结果，那么就是int c = *a + *b; 最后结果是一个数字，不是地址，所以不要...
warning is the “cast from pointer to integer of different size.” This warning occurs when you attempt to cast a pointer type to an integer type that has a different size than the pointer. Understanding this warning is crucial for developers who want to ensure their code is robust and ...
c语言报错cast from pointer to integer of differentsize?char一个字节，unsigned int至少4个字节，把4...
将int变量转为(void*)时出现错误 error: cast to pointer from integer of different size [-Werror=int-to-pointer-cast] 这是由于int类型为32位,指针为long long 64位 解决方法:(voi
modbus.c:861:40: warning: cast from pointer to integer of different size [-Wpointer-to-int-...
main.c:10:18: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast] 主要原因是在源文件中没有声明函数 未经声明的函数原型一律默认为返回int值(4个byte)定义赋值的指针变量(8
今天在编写多线程程序的时候,编译过程中出现了如下错误: thread.c: In function ‘main’: thread.c:38:57: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast] 后来google了,受这个问题解决的启发http://stackoverflow.com/questions/9251102/warning-cast-to-pointer-from-integ...
Therefore, I decided to suppress the error using an intermediate cast to uintptr_t. Summary by CodeRabbit Bug Fixes Resolved priority calculation issues in the queue by adjusting type casting for consistent handling of elements. Suppress error: cast from pointer to integer of different size [-...