you can use combination of if, for and flag
Code:int match = 1; for(k=0;k<strlen(word) && match;k++) { if(!condition_for_current_k) match = 0; } if(match) printf("%s is an anagram of %s\n" dict[i].name, word);
you can use combination of if, for and flag
Code:int match = 1; for(k=0;k<strlen(word) && match;k++) { if(!condition_for_current_k) match = 0; } if(match) printf("%s is an anagram of %s\n" dict[i].name, word);
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler