how to convert decimal to floating point number

Code:

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  mantissa = mantissa << 9;
  mantissa = mantissa >> 9;

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In particular the above seems very pointless...

--
Mats

Code:

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exponent = exponent << 1
exponent = exponent >> 23;

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This is also wrong. Why would you shift the exponent left when you want it at the beginning, to the right? If you remove the left shift, it's correct.

okay I fixed all the errors I have, now how do I translated the mantissa into hexadecimals. like for instance if it's -10 then it should be 0xa00000 in the mantissa

Your bits are in the wrong position, that's why it seems you are getting "excess."
You can use my function for debugging:

Code:

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void PrintBits(uint32_t nNumber)
{
        char Binary[sizeof(nNumber) * 8 + 1] = {0};
        char Binary2[sizeof(nNumber) * 8 + 1] = {0};
        for (int i = sizeof(nNumber) * 8 - 1; i >= 0; i--)
        {
                if (nNumber & (0x1 << i))
                        Binary[i] = '1';
                else
                        Binary[i] = '0';
        }
        for (int i = sizeof(nNumber) * 8 - 1, j = 0; i >= 0; i--, j++) Binary2[j] = Binary[i];
        for (int i = 0; i < sizeof(nNumber) * 8; i++)
        {
                if (i == 1) cout << " ";
                else if (i == 1 + 8) cout << " ";
                cout << Binary2[i];
        }
        cout << "\n";
}

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It will print all bits of a number. Make sure there are no "0" to the right-most, because if there it, you didn't shift it right.
Shift it all the way to the right. And be careful not to shift too much.

Quote:

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Originally Posted by matsp

Code:

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  mantissa = mantissa << 9;
  mantissa = mantissa >> 9;

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In particular the above seems very pointless...

--
Mats

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Depending on the size of the mantissa variable, it could potentially set some of the higher bits to zero. I doubt that is the intention though.

Quote:

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Originally Posted by -EquinoX-

okay I fixed all the errors I have, now how do I translated the mantissa into hexadecimals. like for instance if it's -10 then it should be 0xa00000 in the mantissa

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No, for -10, you should get:

Mantissa 0x00200000
Exponent 0x0082
Sign 0x01

Quote:

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Originally Posted by brewbuck

Depending on the size of the mantissa variable, it could potentially set some of the higher bits to zero. I doubt that is the intention though.

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But only necessary if there wasn't an AND on the line before that already had removed any "unwanted" bits, right?

--
Mats

I think what brewbuck referred to was if you extracted a number such as
1111 11111 0000 0000
Shift left 9 times and we get:
1000 0000 0000 0000
Shift right 9 times and we get:
0000 0001 0000 0000
Unintentionally destroyed data.
But it depends on the size. It it was 32-bit, obviously this wouldn't happen.

yes I am doing this on a 32 bit machine

I mean the size of the variable. My example was of a 16-bit variable.
But it goes to show that you should be careful about shifting the original data since you can unintentionally destroy parts of it.

Quote:

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Originally Posted by Elysia

Your bits are in the wrong position, that's why it seems you are getting "excess."
You can use my function for debugging:

Code:

---

void PrintBits(uint32_t nNumber)
{
        char Binary[sizeof(nNumber) * 8 + 1] = {0};
        char Binary2[sizeof(nNumber) * 8 + 1] = {0};
        for (int i = sizeof(nNumber) * 8 - 1; i >= 0; i--)
        {
                if (nNumber & (0x1 << i))
                        Binary[i] = '1';
                else
                        Binary[i] = '0';
        }
        for (int i = sizeof(nNumber) * 8 - 1, j = 0; i >= 0; i--, j++) Binary2[j] = Binary[i];
        for (int i = 0; i < sizeof(nNumber) * 8; i++)
        {
                if (i == 1) cout << " ";
                else if (i == 1 + 8) cout << " ";
                cout << Binary2[i];
        }
        cout << "\n";
}

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It will print all bits of a number. Make sure there are no "0" to the right-most, because if there it, you didn't shift it right.
Shift it all the way to the right. And be careful not to shift too much.

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which one is wrong?? I am sure that not all is wrong right?

Well, as I mentioned, for the number -10.0, you should get:

Mantissa 0x00200000
Exponent 0x0082
Sign 0x01

If you don't, then there's a bug somewhere in the code. Look at the results you are getting and compare them to these. If they don't match, then they're wrong.

Quote:

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Originally Posted by -EquinoX-

Code:

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  int* sign_pointer = (int*) &number;
  int temp = *sign_pointer;

  printf("%p\n", sign_pointer);

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%p takes a void *.

Code:

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printf("%p\n", (void *) sign_pointer);

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as I said the hidden value, which is 1. x x x x x is included here

my other question is that say that I have an unsigned int 126 and I want to substract that with the int 127 so the result will be -1, how would I do this??

Quote:

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Originally Posted by -EquinoX-

as I said the hidden value, which is 1. x x x x x is included here

my other question is that say that I have an unsigned int 126 and I want to substract that with the int 127 so the result will be -1, how would I do this??

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I usually use "-" to subtract things. Make sure that the result variable is signed. You shouldn't need to cast your unsigned variable, I don't think.