char *数组以及如何为每个数组分配内存

Question

I have a very simple problem that I cannot seem to figure out. I have this:

char* array[10];

So, I then have 10 char* pointers on the stack. Now all I want to do is allocate memory for each pointer. As in:

array[0] = malloc(sizeof(char)*6);

And then store some characters at this location:

strncpy(array[0], "hello", sizeof("hello")); 

Yet, I am getting a compile-time error at the first step of allocating the memory:

error: invalid conversion from ‘void*’ to ‘char*’ [-fpermissive]

But it works as expected at Ideone .

What am I doing wrong? I understand what I am trying to do, but I do not understand why it does not work. At each index in array there is a char* . By using the = symbol I am trying to assign each pointer to a block of memory allocated to it.

What am I doing wrong? Compiling with g++ -g -Wall

Answer 1

What am I doing wrong? Compiling with g++ -g -Wall

g++ always compile a .c file as .cpp . Compile it with a C compiler (like GCC). In C++, you must have to cast the return value of malloc . In case of C, do not cast return value of malloc .

Answer 2

Your code is valid C, but you are compiling your code as C++, which, unike C, has no implicit conversion from void* to char* .

If you intended to compile the code as C (in which case you do not require the cast), use gcc , instead of g++ . Also make sure you your file does not end with an extension that gcc interprets as C++ ( .cpp , .C , .cxx or .cc ). Or play it safe and use the .c extension.

---

If you want to make the code valid C++, you need to cast to char* :

array[0] = (char*)malloc(sizeof(char)*6);

Answer 3

This is probably the most visible difference between C and C++: C can implicitely convert the void* returned by malloc() to any other type, C++ can't.

Now, by compiling with g++ , or by using a .cpp file name extension, you are compiling your code as C++ code, not C code. Use gcc instead and make sure that your source file ends with .c , and your code will compile fine.

An alternative solution is to add the cast that C++ requires: array[0] = static_cast<char*>(malloc(sizeof(char)*6));

Answer 4

As others have pointed out, C++ does not allow an implicit conversion from void * to char * .

If this is really supposed to be C++ code, I'd advise using new instead of malloc for dynamic memory allocation, and for this particular code I'd advise using a vector of string instead of an array of char * :

#include <vector>
#include <string>
...
std::vector< std::string > array;
...
array[0] = "hello";  // literal is implicitly converted to an instance of string

The string and vector implementations do all the memory management for you.

If this is really supposed to be C code, simply compile it using gcc instead of g++ .

Answer 5

Try something like this:

array[0] = static_cast<char *>(malloc(sizeof(char)*6));

How should I cast the result of malloc in C++?