ISO C ++禁止在转换为c ++的旧代码中增加类型为'void *'的指针

Question

i know its old problem , but what will be the easy solution to keep the old logic of c app (legacy) that now converted to c++ ?

in c its working :

void *p;
void *response = malloc(60 * 81);

p = response ;
p+=4;

in g++ gives : ISO C++ forbids incrementing a pointer of type 'void*' update:
if i change it to char* im getting this error:

char *p;
char *response = malloc(60 * 81);

error: invalid conversion from ‘void*’ to ‘char*’

also does char* can hold other types (basic ones ) like short , int , bool ? this is why it is used in this legacy code , to hold diffident types ,

Answer 1

The simplest would be to cast the void * to char * . Since ISO C also forbids void* arithmetic, gcc treats it as a char* as an extension. See this for more details: http://gcc.gnu.org/onlinedocs/gcc-4.7.1/gcc/Pointer-Arith.html#Pointer-Arith

Answer 2

It's only working in gcc. Pointer arthimetic on void * is undefined. Gcc treats it like a char * in that case. So the best way to fix your legacy code is to carefully change all those pointers to char *.

Answer 3

The easiest might be to look for compiler options to change this behavior.

The best is probably to change the type to char * , since that seems to match usage and intent.

Answer 4

Porting from gnu C to C++ is non trivial. Arithmetic on void* is not C, it is an extension.

Porting such things to C++ should be done more carefully, in particular if the C code was not too proper from the start. That data has an "intended" type, so you should use that type in C++ and not yet another second guess like char . Obviously this was not thought to be a C string, doing += 4 for C strings makes not much sense. So there is the assumption that the base type has a size of 4 , probably from the rest of the code you can guess how this has to be interpreted.

Once you have the proper type, use new[] to allocate the array. Don't use malloc in C++ if you can avoid it.

Answer 5

由于“指针的算术”，如果您写p += 4则意味着： p += ((sizeof(void *) * 4)