Integer.parseInt(a+1); parses the String that results from concatenating the value of the String a (“1”) to the int literal 1, which is “11”. Change it to int inc = Integer.parseInt(a) + 1; This way “a” would be parsed to the integer 1 and then 1 would be added to it to give you the value … Read more
def string_gen(length.to_i) chars = %w(a b c d e f g h i j k l m n o p q r s t u w x y z) random_string = chars.to_a[rand(chars.length)].join puts random_string end string_gen(6) #<= “jahsyd” If you want to do it more then one time add a .times loop: def string_gen(length.to_i) chars … Read more
You probably want this: #include <stdio.h> #include <string.h> int main(void) { char temp[256]; char *st[3]; // array of three pointers to char for (int i = 0; i < 3; i++) { scanf(“%255s”, temp); // prevents potential buffer overflow st[i] = strdup(temp); } for(int i = 0; i < 3; i++) { printf(“%s\n”, st[i]); free(st[i]); … Read more
These two questions typically differ based on how they’re framed. Usually, the string permutation question is asked as Given a string w and a string x, check whether w is a permutation of x. In that sense, you know what both strings are, and you just need to see whether the character counts match (or … Read more
Generally, the logic of your code is OK except a return value mistake in the empty() function. In the empty(), it should return 1 while stack is empty. int empty(){ if(top==-1) return(0); <– !!! here should return 1 } Otherwise, it will go into the while loop at while(precedence(value)<=precedence(stack[top])&&!empty()) even when stack is empty. And … Read more
Whenever you need some reasonably common matching problem resolve in Perl, you should always first check the Regexp::Common family on CPAN. In this case: Regexp::Common::Email::Address. From POD Synopsys: use Regexp::Common qw[Email::Address]; use Email::Address; while (<>) { my (@found) = /($RE{Email}{Address})/g; my (@addrs) = map $_->address, Email::Address->parse(“@found”); print “X-Addresses: “, join(“, “, @addrs), “\n”; } 1 … Read more
The amount of downvotes is crazy but oh well… I found the reason for my performance issue / bottleneck thanks to the comments. The second for loop inside StartSimilarityCheck() iterates over all entries, which in itself is no problem but when viewed under performance issues and efficient, is bad. The solution is to only check … Read more
#include <stdio.h> #include <string.h> #include <conio.h> int main(void){ char A[81][81] = {0}; int t=0,j=1,k=0,l; puts(“Input a sentence (max 80 character)”); scanf(“%80[^\n]”, A[0]);//’gets’ has already been abolished, it should explore a different way. while (A[0][t] != ‘\0’){ if(A[0][t] == ‘ ‘){ ++j; k=0; while(A[0][t] == ‘ ‘)//Skip spaces ++t; } else { A[j][k++] = A[0][t++]; } … Read more
I suppose easier way to do this is: size_t len = strlen(buf); // will calculate number of non-0 symbols before first 0 char * newBuf = (char *)malloc(len); // allocate memory for new array, don’t forget to free it later memcpy(newBuf, buf, len); // copy data from old buf to new one 7 solved Remove … Read more
There are two problems here. First, while std::string is null-terminated (required in C++11, de facto in most implementations before that), you cannot access past size(). If you used string::at() directly, then you’d get hit: reference at(size_type pos); Throws: out_of_range if pos >= size() which would be true for the null terminator. So the right way … Read more
In addition to LihOs answer above this block looks wrong: if(tab1[i] != ‘\0′) tab1[i]==result[i]; else if (tab2[i] !=’\0′) tab2[i]==result[i]; else printf(” “); Don’t you mean to assign the value in tab1[i]or tab2[i] to result[i]like this? if(tab1[i] != ‘\0′) result[i] = tab1[i]; else if (tab2[i] !=’\0’) result[i] = tab2[i]; else printf(” “); Also using magic numbers … Read more
As Mofi already noted, a simple for will do the trick. Add a counter to build up a variable for every “word” (array). Whether they are real words or numbers doesn’t matter, but some special chars will generate errors, like &, | or just being ignored, like %, !). @echo off setlocal enabledelayedexpansion set count=0 … Read more
Here’s my solution: #include <stdio.h> int main() { char sentence [100]; int i,letter_count=0; printf(“Enter a sentence: “); fgets(sentence,100,stdin); for(i=0;sentence [i]!=’\0′;i++) { if(sentence [i]==’ ‘ || sentence [i+1]==’\0’) { if(letter_count%2==1) { printf(“%c “,sentence [i-letter_count/2-1]); } letter_count=0; } else { letter_count++; } } putchar(‘\n’); return 0; } 2 solved Can anyone help me how to creat a … Read more
Use String.Replace(): string x = @”documents\bin\debug”; string desiredString = x.Replace(@”\bin\debug”, String.Empty); Note: The key thing here is that you have to assign the string returned by the Replace() function to a variable. (From your comment on the question, it is the problem). This can either be another variable (as in the above example) or the … Read more
Just use array[0].value: var array = [{ label: “1”, value: “11” }, { label: “2”, value: “22” }, { label: “3”, value: “33” }, { label: “4”, value: “44” }, ]; var firstValue = array[0].value; console.log(firstValue); You can also use destructuring like so: var array = [{ label: “1”, value: “11” }, { label: “2”, … Read more