2 chars, one int

ok, little bit of a problem here.

I am reading in chars, x and y. once read, i need to print them out as y, x.

I have an int Z which i need to contain y AND x without adding them. so if x actually was 35 and y actually was 37, i would need Z to store 3735.

once i figure this out i need to shift the int left 4 bits, but i figured i would get this chunk done first.

any help is appreciated.

thanks,
keith

Instead of the 2 chars and 1 int. make 3 chars and use strcat() to put those 2 chars into the 1 char. Ex:

Code:

---

#include <iostream>
#include <cstring>

using namespace std;

int main()
{
        char tempchar1[3];
        char tempchar2[3];
        char totalchars[5] = {"\0"};

        cin >> tempchar1;
        cin >> tempchar2;

        strcat(totalchars, tempchar1);
        strcat(totalchars, tempchar2);
        strcat(totalchars, "\0");

        cout << "Together: " << totalchars << endl;
        return 0;
}

---

hrm. maybe it would help if i posted code.

int a, b;
char xCha, xChb;
int combine;

fIn2.open(userfilename.c_str(), ios::in | ios::binary);
fIn2.seekg(ios::beg);
fIn2.seekp(1000);
while(!fIn2.eof()) {
fIn2.get(xCha);
fIn2.get(xChb);
a= (int)xCha;
b= (int)xChb;
combine = ???;
fout.put(combine);
}

the int "combine" needs to take ints b and a and combine them. for example if "a" reads C3 and "b" reads 23, i would need "combine" to read 23C3.

any help using the code i have?

thanks,
keith

an int is 4 bytes. so you should be able to store your bytes A and B in the low two bytes of your int.

Z = ((int)A<<8) | (int)B;

done.

hrm. i tried both these methods and they don't seem to give me the results i'm looking for. i'm probably just dumb and am missing something basic, but here is what i DO have:

i have two ints.

i need to put these two ints together, without adding them mathematically. like
int a = 12
int b = 78

and int c must = 1278, not 90.

any ideas? is it possible?

thanks again,
keith

our methods did the same thing.

((int)A<<8) | (int)B

gives the same results as

A*256 + B

0x12 * 256 + 0x78 is 0x1278

I wrote a simple program that does what your looking for, i hope it helps

Code:

---

#include <iostream>

using namespace std;

int main()  {

    int a = 12;
    int b = 78;
    int c = 0;
    char buffer[5];

    sprintf(buffer, "%d%d", a, b);
    c = atoi(buffer);
    cout << c << endl;

    system("pause");

return 0;
}

---

hrm. when i use this method:

while(!fIn2.eof()) {
fIn2.get(A);
fIn2.get(B);
combine = ((int)A<<8) | (int)B;
fout.put(combine);

it seems to be printing out every other int, rather than combining both A and B. i'm probably missing something.....also, what does | do in this program?

keith

don't use put. Use write(). I think put just does one byte.

fout.write(&combine,2);

except you should probably understand that this writes out the low two bytes of combine and ignores the high two.

also, since intel is lil-endian, you're bytes will be reverse of what you expect.

"|" operator is the bitwise "OR" meaning if you have the following bytes

01011000
10000110
the result is
11011110

as each bit is or'd

oooh awesome, OR'ing was going to be a question later on!

but i tried using write(&combine,2) and got the following error when compiling:

'write' : cannot convert parameter 1 from 'int *' to 'const char *'
Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast

keith

write((char*)&combine,2)