如何在 for 循环中每次迭代只执行一次“if”语句？

Question

Assuming there are no duplicate words in either list, I would like to compare the words of listA with the words in listB. If there is a match, I want to print the word that matches and compare the next 'n' words in listB to see if there is a match. Likewise, if there is no match, (ie once I reach the last word in listA), I want to print the word that could not be found and compare the next 'n' words in listB to see if there is a match.

I am stuck on how I should implement statements (if, break, continue) in my for loop so that it meets the specifications listed above. When I run the code below, it only prints the instance in which there is a match, but it does not print anything at all if there is no match.

alineno & blineno refer to current line number in the arrays aline & bline where the words are stored

 // index through listA
 for(i = 0; i < alineno; i++){ 
   // index through all the words in listB
   for(j = 0; j < blineno; j++){ 
      if(strcmp(aline[i], bline[j]) == 0){
        printf("%s is in the list!", bline[j]);
      }
      continue;
      if(strcmp(aline[strlen(aline[0])-1], bline[j]) != 0){
        printf("%s is not in the list!", bline[j]);
      }
   }
 }

1. listA: Aardvark,Cat,Bear,Dog
2. listB: Cat,Badger

Cat is in the list! Badger is not in the list!

Cat is in the list!

EDIT:

I understand that my continue statement is the reason why the second condition is not being checked. Removing it would print a word is / is not in the list 'j' amount of times, which is not my desired output. In other words, I would appreciate guidance on how I should implement such statements in order to meet the specifications.

Answer 1

My suggestion is that you change the loops, so you have the loop over "listB" as the outer loop, and iterate over "listA" in the inner loop.

Then you can easily set a flag in the inner loop, and break out of it when a match is found. In the outer loop you check this flag to decide what to print.

In pseudo code perhaps something like this

for (to_find in listB)
{
    found_flag = false;

    for (animal in listA)
    {
        if (to_find == animal)
        {
            found_flag = true;
            break;
        }
    }

    if (found_flag)
        printf("Animal found");
    else
        printf("Animal not found");
}

Answer 2

Your continue is always executed; you will never reach your second if .

Answer 3

First, use goto , like this:

void something(void) {
    // index through listA
    for(int i = 0; i < alineno; i++){ 
        // index through all the words in listB
        for(int j = 0; j < blineno; j++){ 
            if(strcmp(aline[i], bline[j]) == 0){
                printf("%s is in the list!", bline[j]);
                goto doneAnimal;
            }
        }
        printf("%s is not in the list!", bline[i]);
doneAnimal: ;
    }
}

Second; to avoid the risk of "goto is bad" nonsense (see Historical Note below), make the code harder to read by splitting it into 2 different functions, so that you can convert the goto into a return , like this:

void something(void) {
    // index through listA
    for(int i = 0; i < alineno; i++){ 
        doAnimal(i, blineno);
    }
}


void doAnimal(int i, int blineno) {
    for(int j = 0; j < blineno; j++){ 
        if(strcmp(aline[i], bline[j]) == 0){
            printf("%s is in the list!", bline[j]);
            return;
        }
    }
    printf("%s is not in the list!", bline[i]);
}

Historical Note

Once upon a time higher level languages (like assembly language) did not have structured programming features ( do , while , break , continue , switch , ...). Instead programmers would write code using goto , like (eg) " if(x < MAX) goto loopStart; instead of a " } while(x < MAX); .

To encourage the adoption of structured programming features, in 1968 Edsger W. Dijkstra wrote a letter to the editor of ACM entitled "go to statement considered harmful". This letter had the desired effect - structured programming features ( do , while , break , continue , switch , ...) were adopted in all major languages.

However; it also had one unintended side-effect - the letter was a little too effective; and ignorant people (that failed to read the letter or understand its context) started becoming zealots, making their code worse (for cases where the new structured language features aren't enough) to avoid goto without understanding why, and encouraging other people to make their code worse without understanding why.

Examples of this include complicating code by introducing extra variables purely for the sake of avoiding a simpler goto , and/or complicating code to introduce extra branches purely for the sake of avoiding a simpler goto .

Later (in conversations with Donald E. Knuth); Dijkstra himself said " Please don't fall into the trap of believing that I am terribly dogmatical about [the go to statement]. I have the uncomfortable feeling that others are making a religion out of it , as if the conceptual problems of programming could be solved by a single trick, by a simple form of coding discipline! "

Sadly; once ignorance begins to spread common sense is fighting a losing battle.

Answer 4

The best way to do this is probably binary search or a hash table, depending on the amount of data. That being said, the code could be improved in the following way:

for(int i = 0; i < alineno; i++)
{ 
  int j;
  for(j = 0; j < blineno; j++)
  {
    if(strcmp(aline[i], bline[j]) == 0)
      break;
  }

  if(j == blineno)
    printf("%s is not in the list!", aline[i]); 
  else
    printf("%s is in the list!", bline[j]);
}

Note: aline[i] not bline[i] in the printf. bline[i] would be a potential array out of bounds bug, if alineno and blineno are allowed to have different lengths.