CODEWITHSUNDEEP
c++arrayschar
Deniz
Deniz

Reputation: 378

Fill fixed size char array with integers in C++

I want to store integers in char array ex: 0 up to 1449, I checked other posts and I tried memset, sprinf etc. but either I get gibberish characters or unreadable symbols when I print inside of char array. Can anyone help please?

I checked the duplicate link however I am not trying to print int to char, I want to store int in char array. But I tried buf[i] = static_cast<char>(i); inside for loop but it didn't work. Casting didn't work.

The last one I tried is like this: char buf[1450];

for (int i = 0; i < 1449; i++)
{
    memset(buf, ' '+ i, 1450);
    cout << buf[i];
}

The output is: here

Upvotes: 1

Views: 1658

Answers (2)

Frodyne
Frodyne

Reputation: 3973

You are going to have to explain better what it is you want, because "store integers in char array" is exactly what this code does:

char buf[1450];

for (int i = 0; i < 1450; i++)
{
    buf[i] = static_cast<char>(i);
    std::cout << buf[i];
}

Yes, the output is similar to what your picture shows, but that is also the correct output.

When you use a debugger to look at buf after the loop, then it does contain: 0, 1, 2, 3, ..., 126, 127, -128, -127, ..., 0, 1, 2, 3, ... and so on, which is the expected contents given that we are trying to put the numbers 0-1449 into an integer type that (in this case*) can contain the range [-128;127].

If this is not the behavior you are looking for (it sounds like it isn't), then you need to describe your requirements in more detail or we won't be able to help you.

(*) Char must be able to contain a character representative. On many/most systems it is 8 bits, but the size is system dependent and it may also be larger.

New answer.

Thank you for the clarification, I believe that what you need is something like this:

int32_t before = 1093821061; // Int you want to transmit

uint8_t buf[4];
buf[0] = static_cast<uint8_t>((before >> 0) & 0xff);
buf[1] = static_cast<uint8_t>((before >> 8) & 0xff);
buf[2] = static_cast<uint8_t>((before >> 16) & 0xff);
buf[3] = static_cast<uint8_t>((before >> 24) & 0xff);

// Add buf to your UDP packet and send it
// Stuff...
// After receiving the packet on the other end:

int32_t after = 0;
after += buf[0] << 0;
after += buf[1] << 8;
after += buf[2] << 16;
after += buf[3] << 24;

std::cout << before << ", " << after << std::endl;

Your problem (as I see it), is that you want to store 32bit numbers in the 8bit buffers that you need for UDP packets. The way to do this is to pick the larger number apart, and convert it into individual bytes, transmit those bytes, and then assemble the big number again from the bytes.

The above code should allow you to do this. Note that I have changed types to int32_t and uint8_t to ensure that I know the exact size of my types - depending on the library you use, you may have to use plain int and char types, just be aware that then the exact sizes of your types are not guaranteed (most likely it will still be 32 and 8 bits, but they can change in size if you change compiler or compile for a different target system). If you want you can add some sizeof checks to ensure that your types conform to what you expect.

Upvotes: 1

Francisco Basadre
Francisco Basadre

Reputation: 82

I'm not sure what you trying to do! You should say your objective.

A char (usually 8 bit) in c++ doesn't hold an int (usually 32 bit), If you want to store an int you should use an int array:

int buf[1500];

The memset(buf, ' '+ i, 1450); will actually write the sum of ' ' ascii number plus i always at beginning of the buffer (buffer address is never incremented).

something like this, maybe is what you want:

int buf[1500] = 0;
for (int i = 0; i < 1449; i++)
{
    buf[i] = i;
    cout << buf[i] << ' ';
}

consider using c++11 containers like std::vector to hold the int or chars, would much safer to use.

Upvotes: 1

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