当void*转换为int时,会报错: Castfrompointer to smaller type ‘int‘ loses information 后来,发现这么修改即可: (int)(size_t)data
"Cast from pointer to smaller type 'int' loses information” 从错误内容我们也能看出, 错误出在'int'这里了。 而那份代码也比較早的, 在Xcode5.1之后, 要用uintptr_t来替代int。 把错误语句处的int全换成uintptr_t就可以。 可是, 这样可能会带来很多其它的问题。 还有第二种解决的方法: Xcode5.1開始将...
线程池代码上,往往需要输出对应的线程池编号,即(int) pthread_self() 但是在64为机器上可能导致因为int为4字节,指针统统为8字节,所以一般将int改成uintptr_t 即可解决。 即,将(unsigned int)强制类型转换修改成: uintptr_t 即可。
src/annoymodule.cc:182:12: error: cast from pointer to smaller type 'int' loses information return (int) NULL; ^~~~ NULL can't be converted to integer because it is a pointer. In fact, this function call and comparison are meaningl...
Error "Cast from pointer to smaller type 'int' loses information" in EAGLView.mm when update Xcode to 5.1 (5B130a) 10 Disabling "cast from pointer to smaller type uint32_t" error in Clang Related 3 "warning: initialization makes pointer from integer without a cast". Bu...
Im using OS X 10.10.2 Took me quite a while the makefile. And now Im getting this error /include/SYS/SYS_Math.h116812 error cast from pointer to smaller type ‘uint’ (aka ‘unsigned int’) loses information return (uint)ptr;
error: cast from pointer to smaller type 'uintptr_t' (aka 'unsigned int') loses information const auto __intptr = reinterpret_cast<uintptr_t>(__ptr); Also this seems to be a common cast error in c++, Please see if this helps resolve the issuehttps://stackoverflow.com/questions/22419...
到函数里再拿出来就ok了。如果不愿意这样的话，我想你可以传(void *)(long)i进去 ...
编译器报错是 (void *)i 处,错误说明是 Error: cast to 'void *' from smaller integer type 'int' PS: 这是在我上传到远程服务器时构建产生的错误,在我本机UBUNTU使用CLANG3.5编译没有任何问题,我在本机使用的构建命令是 clang -O0 -std=gnu11 -march=native -lm -lpthread pagerank.c -o pagerank...
编辑1：顺便说一句，isnotdigit函数在for循环k != strlen(inp_key)中有一点bug，如果检测到一个非...