It keeps throwing this exact error message even though both types are 32 bits wide. I want to learn the difference between the three type cast operators: Hi, rev2023.5.1.43405. In 64-bit programs, the size of the pointer is 64 bits and it cannot be put into the int type that remains 32-bit almost in all the systems. there How should an excellent flowchart be drawn? Share Improve this answer Follow answered May 6, 2018 at 7:24 Rahul I would like to know the reason. */void **MatrixIB (unsigned long x, unsigned long y, int size){ void *ptr; void **returnPtr; register unsigned long i, j; returnPtr = (void **) malloc (x * sizeof(void *)); ptr = (void *) malloc (x * y * size); for (i=0, j=0; j returnPtr[j] = (void *) (ptr + i); // <<< Compile Errors, Error1error C2036: 'void *' : unknown sizec:\temp\testone\lib\utils.c57, 2> returnPtr[j] = (void *) ((long*)ptr + i); // <<< No compile errors, 3> returnPtr[j] = (void *) ((char*)ptr + i); // <<< No compile errors. for saving RAM), use union, and make sure, if the mem address is treated as an int only if you know it was last set as an int. Making statements based on opinion; back them up with references or personal experience. The OP wanted to convert a pointer value to a int value, instead, most the answers, one way or the other, tried to wrongly convert the content of arg points to to a int value. Losing bytes like thisis called 'truncation', and that's what the first warning is telling you. Two MacBook Pro with same model number (A1286) but different year. I don't understand why the following two cases produce different Hi, I've this question: suppose we have two differently typed pointers: long guarantees a pointer size on Linux on any machine. this way you won't get any warning. static_cast conversion - cppreference.com He also rips off an arm to use as a sword. I need to convert the argument to an int for later use: The compiler (GCC version 4.2.4) returns the error: You can cast it to an intptr_t type.
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