如何在 C 中编写从 a 到 z 和 A 到 Z 运行的单个 for 循环？

Question

I want to combine both the for loops into single for loop. How can i do that?

I want to loop through a to z, and A to Z, like so:

char ch;
for (ch = 'A' ; ch <= 'Z' ; ch++ )
{ 
}
for (ch = 'a' ; ch <= 'z' ; ch++ )
{
}

but using a single loop.

Answer 1

for (char ch = 'A' ; ch <= 'z' ; ch == 'Z' ? ch = 'a' : ++ch )
{
}

Should work -- though please, please, don't inflict this on your fellow developers.

Answer 2

I don't personally like this solution, but:

char * letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
for (char * ptr = letters; *ptr != 0; ++ptr) {
    char ch = *ptr;
    ...
} 

Answer 3

You can do it in a nested loop (two loops, but only one body):

for (start = 'A'; start <= 'a'; start += 'a' - 'A') {
    end = start + 'Z' - 'A';
    for (ch = start; ch <= end; ++ch) {
         /* body */
    }
}

Answer 4

Well, the obvious question is why? ...and the second question is do you care about non-ASCII character sets (as your two loops will fail for EBCDIC), but the quick and dirty way of connecting the two is

for (ch = 'A'; ch <= 'z'; ch++) {
    if (ch > 'Z' && ch < 'a') ch = 'a';
     :

Answer 5

const char diff = 'a' - 'A';
for (ch = 'A' ; ch <= 'Z' ; ch++ )
{
  char small_ch = ch + diff;
 //... 
}

Answer 6

Try this:

        for (int i = 0; i < 52; ++i)
            printf("%c\n", 'A' + i + ('a' - 'Z' - 1) * (i/26));

Answer 7

for (int i=0;i<26;++i) {

     char ch = 'A' + i;
     // Logic for Uppercase letters.

     char ch1 = 'a' + i;
     // Logic for Lowercase letters.

}

Answer 8

#include<stdio.h>
#include<conio.h>
int main()
{
    int i;
    char e;
    for(i=65;i<=122;i++)
    {
                        if(i<91||i>96)
                        {
                            e=i;
                        printf("%c\n",e);
                        }
                        }

    getch();
}

Answer 9

A straightforward solution is

int i;
for(i = 0; i < 52; i++){
  char ch = i + (i < 26? 'A' : 'a');
  /* do something with ch */
}

although I prefer, especially in sensible languages that allow nested functions,

for(ch = 'A'; ch <= 'Z'; ch++)
  dosomething(ch);
for(ch = 'a'; ch <= 'z'; ch++)
  dosomething(ch);

PS Kobe, I see in one of your comments that your reason for the loops is to check whether a character is a letter... but looping is a horrid way to do that. You could simply do

if(('A' <= c && c <= 'Z') || ('a' <= c && c <= 'z')){
    /* c is a letter */
}

or, considerably better,

#include ctype.h
...
if(isalpha((unsigned char)c)){
    /* c is a letter */
}

(To understand why that cast is needed, read the isalpha man page and the C language standard. This is one of several abominable aspects of C.)

Answer 10

for (char ch = 'A'; ch <= 'z'; ch = ch == 'Z'?'a':ch+1) {
       //loop body
}

This approach is similar to Billy's, but with a slightly less nasty loop-increment statement. I wouldn't mind inflicting this on a fellow dev, though I might write the increment statement as a function to clarify if the increment statement got any more complex:

char nextChar(char c) { return c == 'Z' ? 'a' : c+1; }

for (char ch = 'A'; ch <= 'z'; ch = nextChar(ch)) {
       //loop body
}

Answer 11

It's pointless, just use one for loop. The variables are mainly used as counters, or boundaries for the loop so it knows when to terminate. Just make one loop. Two loops are completely unnecessary.

Answer 12

void main()
{   
    char ch;

    for(ch= 65 ; ch <= 122 ; ch++)
    {
        if ( ch>=91 && ch<=96)
        continue;                     //hey check this out if this helps
        printf("%c \n",ch);
    }

}

Answer 13

for (ch = 'A';ch <= 'z';ch++) {

}

This will also include a few characters such as '[' and ']'...