• C

How to concatenate ints in C?

My program runs rand(), gets it as 16 bit,

eg rand() gives me 12F3
another rand() gives me 564A

so if I do printf("%X%X\n", rand() rand());
I get exactly what I want, 32 bit numbers.

The thing is, I want this in a variable, so I tried to be creative and did:

x = printf("%X%X\n", rand() rand())

but unfortunately didnt work.

So I tried

int x;
int y;
int z;

x = rand()
y = rand()

strcat(x, y);

But C wont accept mixing integer types, now what is my option?

Thanks in advance,

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Hi markov123,

snprintf (buffer, BUFFER_LENGTH, "%X%X",rand(),rand());

http://ddas77.tripod.com/guide/programming/c/sprintf.html     C fundas! the sprintf( ) function...

snprintf --> sprintf

Hi callrs,

snprintf is safer version of sprintf where you can prevent buffer overruns by specifying the max length (buffer capacity) of string to be written to buffer.

ok nvm, it seems snprintf is a function too
markov123Author Commented:
Hi Sunnycoder,

Am still not getting the results I want, I must be doing something wrong.

Please check the syntax below.

   int x;
   int y;
   int z[2];
   x = 0X1111;
   y = 0X3222;
   printf("x: %0x, y: %0x\n", x, y);

   sprintf(z, "%0x%0x\n", x, y);
   printf("%0x\n", z);

z should display 11113222, but instead i get something entirely different...
z should be a char array or char * ... It will be printed using %s and not %x format specifier .. it will be a string ...

s -> string
n -> number of characters
print -> print
f -> formatted
  char z[64];
   sprintf(z, "%0x%0x\n", x, y);
   printf("%s", z);
I think what you're looking for is :

    unsigned int x = ((rand() << 16) & 0xFFFF0000) + (rand() & 0x0000FFFF);

Is that what you need ?

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markov123Author Commented:

This is exactly what I wanted, thanks a lot.


I want z as an int, so when I did: printf("%0xz\n", (int) z); I was getting different results.

Anyway, its working fine now, I thank you both for your efforts.

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