warning: passing argument 1 of ‘memcpy’ makes pointer from integer without a cast

Hey I'm getting this warning and when i run the program the warning actually leads to segementation fault.

im trying to memcpy(unsigned int variable, (void*) buffer, 4);

but I cant seem to get the variable to lose that warning sign

Can you post the code including the headers

You should be passing a pointer as the first parameter, a pointer to the destination you wish to fill.

Code:

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      #include <string.h>

      void *memcpy(void *dest, const void *src, size_t n);

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See the man page for details. What are you trying to do? Just copy variable into buffer? If so, you can just use an assignment . . . memcpy() is intended for when you wish to copy one array into another, or something similar.

well im trying to copy a unsigned int with 4 bytes of the buffer so how do i get about doing that since the unsigned int is 4 bytes and the buffer has like 10 bytes how do i get 4 bytes from buffer into the int??
unsigned char buffer[10]

Just like dwks said - pass the address of the int, not the int itself.

Quote:

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Originally Posted by Dino

Just like dwks said - pass the address of the int, not the int itself.

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Actually, that might not work as expected. Most machines are little endian, but there are some big endian machines. Meaning that a buffer \x01\x02\x03\x04 will usually become 0x04030201, but on other machines 0x01020304. Not to talk about the fact an integer is not even always 4 bytes.
Strictly speaking, there is no way in doing this. Strictly speaking I don't believe you can even do anything completely portable in C. Imagine a 1 bit byte, 1 byte char, short, int and 1 byte of memory. Isn't that allowed according to the standards, as long as the character set only contains 2 characters..? Well, I'm not entirely sure, but I've never read a minimum width of a datatype, except int >= short >= char, char must be able to represent any character in the basic character set.
Okay, let me switch off my pedantic mode for a bit.

To properly do this, make sure that your data type is at least 4 bytes. I'd use uint32_t, even though it's not standards compliant (at least in C++, is it in C?) but even if it doesn't exist you should be able to add your own typedefs. Then, do something like this:

Code:

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var = (p[0] << 24) | (p[1] << 16) | (p[2] << 8) | (p[3] << 0);

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Yes, I do know bit shifting to the left 0 positions doesn't have any effect, but I find the intentions more readable if we do it like this. It'll be optimized out anyway.

uint32_t is part of the C99 standard. uint32_t(3) - Linux man page

Quote:

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Strictly speaking I don't believe you can even do anything completely portable in C.

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Hmm, well, that's a pretty broad statement . . . .

Anyway, this might be more complicated than it needs to be. Where do you get the data in buffer? If you're reading it from a file, for example, then you can use fread() to read an int directly. (Of course, that ignores the endianness issue, so maybe this way is better.)

Another suggestion:

Code:

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uint32_t i = 0;
char buffer[] = {1, 2, 3, 4};
int byte = 0;
while(byte < sizeof(i)) {
    i |= buffer[byte] << ((sizeof(i) - byte - 1) * CHAR_BIT);
}

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Never mind, that turned out to be way more complicated than it needed to be.

okay thanks but I figured out that this could work but iono if theres a problem
unsigned int variable
memcpy((unsigned char*)& variable, (void*) buffer, 4);

memcpy(&variable, buffer, 4);
No need for casts.
Also, though not standard, uint32_t exists in boost, a widely popular C++ library, making it rather portable.