Outputting a-z as numbers (a=0...)

[cplusplusnoob]

I have looked in my programing book and on google and have found no solution to my problem.I am trying to create a program in c++ that the letters a-z using a loop a=0 and adding 1 each time so it should look like this
a 0
b 1
-------
z 25
when i try this i dont know if i get it right but i get a buch of rancom symbols like a heart or spade.please help.

[Scoody]

As you can see from the ASCII-table, lowercase 'a' starts at Dec = 97. So your loop has to start at 97, not 0.
For example:

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for(int i = 0; i <= 25; i++)
  cout << static_cast<char>(i+97) << endl;

[cplusplusnoob]

thank you so much.

[wtetzner]

As you can see from the ASCII-table, lowercase 'a' starts at Dec = 97. So your loop has to start at 97, not 0.
For example:

Code: Select all

for(int i = 0; i <= 25; i++)
  cout << static_cast<char>(i+97) << endl;

This is usually considered better practice, as it makes your code more readable.

Code: Select all

for(int i = 0; i < 26; i++)
  cout << static_cast<char>('a' + i) << endl;

That way, just by looking at the code you can tell it's looping through the lower case letters of the alphabet.

[Ginto8]

As you can see from the ASCII-table, lowercase 'a' starts at Dec = 97. So your loop has to start at 97, not 0.
For example:

Code: Select all

for(int i = 0; i <= 25; i++)
  cout << static_cast<char>(i+97) << endl;

This is usually considered better practice, as it makes your code more readable.

Code: Select all

for(int i = 0; i < 26; i++)
  cout << static_cast<char>('a' + i) << endl;

That way, just by looking at the code you can tell it's looping through the lower case letters of the alphabet.

No need to static_cast it methinks. I'm pretty sure you can just do this (made i++ into ++i, 'cuz I feel like it):

Code: Select all

for(int i = 0; i < 26; ++i)
    cout << (char)((int)'a' + i) << endl;

[wtetzner]

No need to static_cast it methinks. I'm pretty sure you can just do this (made i++ into ++i, 'cuz I feel like it):

Code: Select all

for(int i = 0; i < 26; ++i)
    cout << (char)((int)'a' + i) << endl;

Yeah. You actually don't even need the (int) cast.

Code: Select all

for(int i = 0; i < 26; ++i)
    cout << (char)('a' + i) << endl;

[Scoody]

I try to stay pure C++ and therefore, static_cast.

[wtetzner]

I try to stay pure C++ and therefore, static_cast.

Yeah, that's fine. Although C++ is a superset of C, so technically any C code is still "pure" C++.

[cplusplusnoob]

As you can see from the ASCII-table, lowercase 'a' starts at Dec = 97. So your loop has to start at 97, not 0.
For example:

Code: Select all

for(int i = 0; i <= 25; i++)
  cout << static_cast<char>(i+97) << endl;

this does work but i am not sure this was the lesson that the book was trying to teach me i will give it to you just as the book gives it to me and again thank you for your time.
"
try this
the character 'b' is char('a'+1), 'c' is cahr(a+2), etc. Use a loop to write out a table of characters with thier corresponding integer values:
a 97
b 98
...
z 122
"

[Scoody]

this does work but i am not sure this was the lesson that the book was trying to teach me i will give it to you just as the book gives it to me and again thank you for your time.
"
try this
the character 'b' is char('a'+1), 'c' is cahr(a+2), etc. Use a loop to write out a table of characters with thier corresponding integer values:
a 97
b 98
...
z 122
"

It's almost the same, the difference is that your book is using "C-style casts", they give you more power, which can be good or bad.
Going along the lines of your book you could do it like this:

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  for(int i = 0; i <= 25; i++)
    cout << char('a' + i) << " " << 'a'+i << endl;

So when i=1 then 'a'+1 would be 98 and when casting it to a char you get 'b'. And since the last 'a'+1 isn't casted you get the corresponding ASCII number.

[MarauderIIC]

Yeah, Scoody's the only person I've seen actually use C++ casts so far. :)

[MarauderIIC]

I split this from the Intro to Programming sticky.

[cplusplusnoob]

dont think my last post made anyway i get it thank you so much falco was really right about the forums

[programmerinprogress]

I guess if you wanted to make the Ascii code of 'A' (or 'a') produce a number equal to one, you good do a bit of bit manipulation .

The Ascii value for 'A' is 65 (off the top of my head), which is 0100 0001 in binary.

So if you get 'A', store it as an integer, and then AND (Represented by a single apersand in C/C++ '&') it by 0011 1111 (or 3F in hex), you should be able to 'knock' the 64 off, to make the resultant value correspondant to the position of the letter in the Alphabet

you could write a program like this

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#include <iostream>
#include <cctype>

using namespace std;

int main()
{
    char Letter;
    cin >> Letter; // get a character from keyboard
    int Value(0);
    Value = Letter; // put ASCII value into an integer

    if(isupper(Letter))//checks for uppercase
    {
        Value = Value & 0x3F;
        //AND the bitfield to change the value
    }
    else
    {
        Value = Value & 0x1F;
        //AND the bitfield to change the value
    }

    cout << Value; // show position in alphabet
    return 0;
}

I know the priorities of this post changed a little, due to the fact that the poster was probably more interested in getting the ascii value, but I thought I would demonstrate this anyway :D