十进制字符串到字符的ASCII转换-C

Question

Can someone explain how to convert a string of decimal values from ASCII table to its character 'representation' in C ? For example: user input could be 097 and the function would print 'a' on the screen, but also user could type in '097100101' and the function would have to print 'ade' etc. I have written something clunky that does the opposite operation:

char word[30];
scanf("%s", word);

while(word[i]!=0)
{
    if(word[i]<'d')
        printf("0%d", (int)word[i]);
    if(word[i]>='d')
        printf("%d", (int)word[i]);
    i++;
}

but it works. Now I want to have function that works in a similar way but of course does decimal > char conversion. The point is, I cannot use any functions like 'atoi' or something like that (not sure about names, never used them ;)).

Answer 1

You can use this function instead of atoi:

char a3toc(const char *ptr)
{
    return (ptr[0]-'0')*100 + (ptr[1]-'0')*10 + (ptr[0]-'0');
}

So, a3toc("102") will return the same thing as (char) 102 , which is an 'f' .

If you don't see why, substitute in the values: ptr[0] is '1' , so the first part becomes ('1'-'0')*100 or 1*100 or 100 , which is what that first 1 in 102 represents.

Answer 2

Tokenize the input string. I'm assuming you are forcing that every letter MUST be represented in 3 characters. So break the string that way. And simply use explicit type casting to get the desired character.

I don't think I should be giving you the code for this, since it is pretty easy and seems more like a Homework question.