Is there a shorter way to write compound 'if' conditions? [duplicate]

Answer 1

If you need to check a character against an arbitrary set of characters, you could try writing this:

std::set<char> allowed_chars = {'A', 'B', 'C', 'D', 'E', 'G', 'Q', '7', 'z'};
if(allowed_chars.find(ch) != allowed_chars.end()) {
    /*...*/
}

Answer 2

Yet another answer on this overly-answered question, which I'm just including for completeness. Between all of the answers here you should find something that works in your application.

So another option is a lookup table:

// On initialization:
bool isAcceptable[256] = { false };
isAcceptable[(unsigned char)'A'] = true;
isAcceptable[(unsigned char)'B'] = true;
isAcceptable[(unsigned char)'C'] = true;

// When you want to check:
char c = ...;
if (isAcceptable[(unsigned char)c]) {
   // it's 'A', 'B', or 'C'.
}

Scoff at the C-style static casts if you must, but they do get the job done. I suppose you could use an std::vector<bool> if arrays keep you up at night. You can also use types besides bool. But you get the idea.

Obviously this becomes cumbersome with e.g. wchar_t, and virtually unusable with multibyte encodings. But for your char example, or for anything that lends itself to a lookup table, it'll do. YMMV.

Answer 3

There is solution to your problem, not in language but in coding practices - Refactoring.

I'm quite sure that readers will find this answer very unorthodox, but - Refactoring can, and is used often to, hide a messy piece of code behind a method call. That method can be cleaned later or it can be left as it is.

You can create the following method:

private bool characterIsValid(char ch) {
    return (ch == 'A' || ch == 'B' || ch == 'C' || ..... );
}

and then this method can be called in a short form as:

if (characterIsValid(ch)) ...

Reuse that method with so many checks and only returning a boolean, anywhere.

Answer 4

In this case you could use a switch:

switch (ch) {
case 'A':
case 'B':
case 'C':
    // do something
    break;
case 'D':
case 'E':
case 'F':
    // do something else
    break;
...
}

While this is slightly more verbose than using strchr, it doesn't involve any function calls. It also works for both C and C++.

Note that the alternate syntax you suggested won't work as you might expect because of the use of the comma operator:

if  ( ch == 'A', 'B', 'C', 'D', 'E', 'F' )

This first compares ch to 'A' and then discards the result. Then 'B' is evaluated and discarded, then 'C', and so forth until 'F' is evaluated. Then 'F' becomes the value of the conditional. Since any non-zero value evaluated to true in a boolean context (and 'F' is non-zero), then the above expression will always be true.

Answer 5

For a simple and effective solution, you can use memchr():

#include <string.h>

const char list[] = "ABCXZ";
if (memchr(list, ch, sizeof(list) - 1)) {
    // 'ch' is 'A', 'B', 'C', 'X', or 'Z'
}

Note that memchr() is better suited than strchr() for this task as strchr() would find the null character '\0' at the end of the string, which is incorrect for most cases.

If the list is dynamic or external and its length is not provided, the strchr() approach is better, but you should check if ch is different from 0 as strchr() would find it at the end of the string:

#include <string.h>

extern char list[];
if (ch && strchr(list, ch)) {
    // 'ch' is one of the characters in the list
}

Another more efficient but less terse C99 specific solution uses an array:

#include <limits.h>

const char list[UCHAR_MAX + 1] = { ['A'] = 1, ['B'] = 1, ['C'] = 1, ['X'] = 1, ['Z'] = 1 };
if (list[(unsigned char)ch]) {
    /* ch is one of the matching characters */
}

Note however that all of the above solutions assume ch to have char type. If ch has a different type, they would accept false positives. Here is how to fix this:

#include <string.h>

extern char list[];
if (ch == (char)ch && ch && strchr(list, ch)) {
    // 'ch' is one of the characters in the list
}

Furthermore, beware of pitfalls if you are comparing unsigned char values:

unsigned char ch = 0xFF;
if (ch == '\xFF') {
    /* test fails if `char` is signed by default */
}
if (memchr("\xFF", ch, 1)) {
    /* test succeeds in all cases, is this OK? */
}

Answer 6

The X-Y answer on the vast majority of modern systems is don't bother.

You can take advantage of the fact that practically every character encoding used today stores the alphabet in one sequentially-ordered contiguous block. A is followed by B, B is followed by C, etc... on to Z. This allows you to do simple math tricks on letters to convert the letter to a number. For example the letter C minus the letter A , 'C' - 'A', is 2, the distance between c and a.

Some character sets, EBCDIC was discussed in the comments above, are not sequential or contiguous for reasons that are out of scope for discussion here. They are rare, but occasionally you will find one. When you do... Well, most of the other answers here provide suitable solutions.

We can use this to make a mapping of letter values to letters with a simple array:

//                    a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p, q,r,s,t,u,v,w,x,y, z
int lettervalues[] = {1,3,3,2,1,4,2,4,1,8,5,1,3,1,1,3,10,1,1,1,1,4,4,8,4,10};

So 'c' - 'a' is 2 and lettervalues[2] will result in 3, the letter value of C.

No if statements or conditional logic required what-so-ever. All the debugging you need to do is proof reading lettervalues to make sure you entered the correct values.

As you study more in C++, you will learn that lettervalues should be static (current translation unit-only access) and const (cannot be changed), possibly constexpr (cannot be changed and fixed at compile time). If you don't know what I'm talking about, don't worry. You'll cover all three later. If not, google them. All are very useful tools.

Using this array could be as simple as

int ComputeWordScore(std::string in)
{
    int score = 0;
    for (char ch: in) // for all characters in string
    {
        score += lettervalues[ch - 'a'];
    }
    return score;
}

But this has two fatal blind spots:

The first is capital letters. Sorry Ayn Rand, but 'A' is not 'a', and 'A'-'a' is not zero. This can be solved by using std::tolower or std::toupper to convert all input to a known case.

int ComputeWordScore(std::string in)
{
    int score = 0;
    for (char ch: in) // for all characters in string
    {
        score += lettervalues[std::tolower(ch) - 'a'];
    }
    return score;
}

The other is input of characters that aren't letters. For example, '1'. 'a' - '1' will result in an array index that is not in the array. This is bad. If you're lucky your program will crash, but anything could happen, including looking as though your program works. Read up on Undefined Behaviour for more.

Fortunately this also has a simple fix: Only compute the score for good input. You can test for valid alphabet characters with std::isalpha.

int ComputeWordScore(std::string in)
{
    int score = 0;
    for (char ch: in) // for all characters in string
    {
        if (std::isalpha(ch))
        {
            score += lettervalues[std::tolower(ch) - 'a'];
        }
        else
        {
            // do something that makes sense here. 
        }
    }
    return score;
}

My something else would be return -1;. -1 is an impossible word score, so anyone who calls ComputeWordScore can test for -1 and reject the user's input. What they do with it is not ComputeWordScore's problem. Generally the stupider you can make a function, the better, and errors should be handled by the closest piece of code that has all the information needed to make a decision. In this case, whatever read in the string would likely be tasked with deciding what to do with bad strings and ComputeWordScore can keep on computing word scores.