My test coverage is up to 66%, how can I increase?
I do not know how to acces or use the alerta variable into the test class, I want to increase the test covergae from 66% to atleast 75% or even more
Apex class:
public static Boolean getContacto (String contactId) {
Boolean alerta = false;
if(String.isNotBlank(contactId) && String.isNotEmpty(contactId)){
Contact contacto =[SELECT Id, UniqueId_Status__c, UniqueId__c from Contact where Id =: contactId and Country__c = 'CHILE'];
if(contacto.UniqueId__c == null && (contacto.UniqueId_Status__c == 'M2' || contacto.UniqueId_Status__c == 'M3' || contacto.UniqueId_Status__c == 'M4' || contacto.UniqueId_Status__c == 'R2' || contacto.UniqueId_Status__c == 'R3' || contacto.UniqueId_Status__c == 'R4')){
alerta = true;
}else
if(contacto.UniqueId__c != null && (contacto.UniqueId_Status__c == 'M2' || contacto.UniqueId_Status__c == 'M3' || contacto.UniqueId_Status__c == 'M4' || contacto.UniqueId_Status__c == 'R2' || contacto.UniqueId_Status__c == 'R3' || contacto.UniqueId_Status__c == 'R4')){
alerta = true;
}else
if(contacto.UniqueId__c != null && contacto.UniqueId_Status__c != 'M2' && contacto.UniqueId_Status__c != 'M3' && contacto.UniqueId_Status__c != 'M4' && contacto.UniqueId_Status__c != 'R2' && contacto.UniqueId_Status__c != 'R3' && contacto.UniqueId_Status__c != 'R4' && contacto.UniqueId_Status__c != 'D1' && contacto.UniqueId_Status__c != null){
alerta = true;
}
}
return alerta;
}
TestClass:
#isTest public class ATCL_BloquearContactoTest { #IsTest
static void bloqueaContactoTest(){
Contact contactos = new Contact();
contactos.FirstName = 'Amanda';
contactos.LastName = 'testing';
contactos.UniqueId_Status__c = 'M1';
contactos.Country__c = 'CHILE';
insert contactos;
Test.startTest();
ATCL_BloquearContactoCTRL.getContacto(contactos.Id);
Test.stopTest();
} }
I do not Know how to coverage alerta variable
Thanks Juan
The low key is to reckon those ifs as diferent scenarios.
To achive those diferente scenarios you must add the new values into the test class to eventually pass the ifs sentences. Every scenrio should be resolve with a difrent #istest method:
public class ATCL_BloquearContactoTest {
#IsTest
static void bloqueaContactoTestNull(){
Contact contactos = new Contact();
contactos.FirstName = 'Amanda';
contactos.LastName = 'testing';
contactos.UniqueId__c = null;//== null
contactos.UniqueId_Status__c = 'M2';//like M_ || R_
contactos.Country__c = 'CHILE';
insert contactos;
Test.startTest();
ATCL_BloquearContactoCTRL.getContacto(contactos.Id);
Test.stopTest();
}
#IsTest
static void bloqueaContactoTestNotNull(){
Contact contactos = new Contact();
contactos.FirstName = 'Amanda';
contactos.LastName = 'testing';
contactos.UniqueId__c = '21354689';//!= null
contactos.UniqueId_Status__c = 'M2';//like M_ || R_
contactos.Country__c = 'CHILE';
insert contactos;
Test.startTest();
ATCL_BloquearContactoCTRL.getContacto(contactos.Id);
Test.stopTest();
}
#IsTest
static void bloqueaContactoTestNotNullNotNull(){
Contact contactos = new Contact();
contactos.FirstName = 'Amanda';
contactos.LastName = 'testing';
contactos.UniqueId__c = '21354689';//!= null
contactos.UniqueId_Status__c = '0';//!(like M_ || R_)
contactos.Country__c = 'CHILE';
insert contactos;
Test.startTest();
ATCL_BloquearContactoCTRL.getContacto(contactos.Id);
Test.stopTest();
}
}
Bye
Related
I want to iterate char by char in a vector of strings. In my code I created a nested loop to iterate over the string, but somehow I get an out of range vector.
void splitVowFromCons(std::vector<std::string>& userData, std::vector<std::string>& allCons, std::vector<std::string>& allVows){
for ( int q = 0; q < userData.size(); q++){
std::string userDataCheck = userData.at(q);
for ( int r = 0; r < userDataCheck.size(); r++){
if ((userDataCheck.at(r) == 'a') || (userDataCheck.at(r) == 'A') || (userDataCheck.at(r) == 'e') || (userDataCheck.at(r) == 'E') || (userDataCheck.at(r) == 'i') || (userDataCheck.at(r) == 'I') || (userDataCheck.at(r) == 'o') || (userDataCheck.at(r) == 'O') || (userDataCheck.at(r) == 'u') || (userDataCheck.at(r) == 'U')){
allVows.push_back(userData.at(r));
}
else if ((userDataCheck.at(r) >= 'A' && userDataCheck.at(r) <= 'Z') || (userDataCheck.at(r) >= 'a' && userDataCheck.at(r) <= 'z')){
allCons.push_back(userData.at(r));
}
else {
continue;;
}
}
}
}
The error here is in these lines:
allVows.push_back(userData.at(r));
allCons.push_back(userData.at(r));
the r variable is your index into the current string, but here you're using it to index into the vector, which looks like a typo to me. You can make this less error prone using range-for loops:
for (const std::string& str : userData) {
for (char c : str) {
if (c == 'a' || c == 'A' || ...) {
allVows.push_back(c);
}
else if (...) {
....
}
}
}
which I hope you'll agree also has the benefit of being more readable due to less noise. You can further simplify your checks with a few standard library functions:
for (const std::string& str : userData) {
for (char c : str) {
if (!std::isalpha(c)) continue; // skip non-alphabetical
char cap = std::toupper(c); // capitalise the char
if (cap == 'A' || cap == 'E' || cap == 'I' || cap == 'O' || cap == 'U') {
allVows.push_back(c);
}
else {
allCons.push_back(c);
}
}
}
Since this question is about debugging actually, I think it is a nice illustration of how the usage of std::algorithms of C++ can decrease the effort needed to see what is wrong with a non working code.
Here is how it can be restructured:
bool isVowel(char letter)
{
return letter == 'A' || letter == 'a' ||
letter == 'E' || letter == 'e'||
letter == 'O' || letter == 'o'||
letter == 'Y' || letter == 'y'||
letter == 'U' || letter == 'u';
}
bool isConsonant(char letter)
{
return std::isalpha(letter) && !isVowel(letter);
}
void categorizeLetters(const std::vector<std::string> &words, std::vector<char> &vowels, std::vector<char> &consonants)
{
for( const std::string &word : words){
std::copy_if(word.begin(), word.end(), std::back_inserter(vowels), isVowel);
std::copy_if(word.begin(), word.end(), std::back_inserter(consonants), isConsonant);
}
}
With a solution like this, you avoid the error-prone access-with-index that lead to your problem. Also, code is readable and comprehensive
I am new to programming and I need to search any string to see if it includes only the letters a,b,c,d,e or f. The minute the program finds a letter that is not one of those the program should return false. Here is my function
bool is_favorite(string word){
int length = word.length(); // "word" is the string.
int index = 0;
while (index < length) {
if ((word[index] == 'a') || (word[index] == 'b') || (word[index] == 'c')||
(word[index] == 'd')|| (word[index] == 'e')|| (word[index] == 'f')) {
return true;
}
else {
return false;
}
index++;
}
}
Thank you very much for nay help! :)
The moment the return statement is encountered, the function is exited. This means that the moment any of the characters 'a', 'b', 'c', 'd', 'e', 'f' is encountered while iterating, due to the return statement the function will be exited immediately.
You can use std::string::find_first_not_of as shown below:
std::string input = "somearbitrarystring";
std::string validChars = "abcdef";
std::size_t found = input.find_first_not_of(validChars);
if(found != std::string::npos)
std::cout << "Found nonfavorite character " <<input[found]<<" at position "<<found<< std::endl;
else
{
std::cout<<"Only favorite characters found"<<std::endl;
}
If you unroll the loop by hand, you will spot the problem immediately:
if ((word[0] == 'a') || (word[0] == 'b') || (word[0] == 'c')||
(word[0] == 'd')|| (word[0] == 'e')|| (word[0] == 'f')) {
return true;
}
else {
return false;
}
if ((word[1] == 'a') || (word[1] == 'b') || (word[1] == 'c')||
(word[1] == 'd')|| (word[1] == 'e')|| (word[1] == 'f')) {
return true;
}
else {
return false;
}
//...
That is, the return value depends only on the first element.
"The minute the program finds a letter that is not one of those the program should return false" means
if ((word[0] != 'a') || (word[0] != 'b') || (word[0] != 'c')||
(word[0] != 'd')|| (word[0] != 'e')|| (word[0] != 'f')) {
return false;
}
if ((word[1] != 'a') || (word[1] != 'b') || (word[1] != 'c')||
(word[1] != 'd')|| (word[1] != 'e')|| (word[1] != 'f')) {
return false;
}
// ...
// After checking all the characters, you know what all them were in
// your desired set, so you can return unconditionally.
return true;
or, with a loop:
while (index < length) {
if ((word[index] != 'a') || (word[index] != 'b') || (word[index] != 'c')||
(word[index] != 'd')|| (word[index] != 'e')|| (word[index] != 'f')) {
return false;
}
index++;
}
return true;
bool is_favorite(string word){
return ( word.find_first_not_of( "abcdef" ) == std::string::npos );
}
It returns true if, and only if, there are only the characters 'a' through 'f' in the string. Any other character ends the search immediately.
And if you exchange string word with const string & word, your function will not have to create a copy of each word you pass to it, but work on a read-only reference to it, improving efficiency.
bool is_favorite(string word){
int length = word.length(); // "word" is the string.
int index = 0;
while (index < length) {
if (word[index] > 'f' || word[index] < 'a')
return false;
index++;
}
return true;
}
The return true is logically in the wrong place in your code.
Your version returns true as soon as it finds one letter that is a through f. It's premature to conclude that the whole string is valid at that point, because there may yet be an invalid character later in the string.
bool is_favorite(string word){
int length = word.length(); // "word" is the string.
int index = 0;
while (index < length) {
if ((word[index] == 'a') || (word[index] == 'b') || (word[index] == 'c')||
(word[index] == 'd')|| (word[index] == 'e')|| (word[index] == 'f')) {
return true; // This is premature.
}
else {
return false;
}
index++;
}
}
Minimal change that illustrates where the return true should be: after the loop. The return true is reached only if and only if we did not detect any invalid characters in the loop.
bool is_favorite(string word){
int length = word.length(); // "word" is the string.
int index = 0;
while (index < length) {
if ((word[index] == 'a') || (word[index] == 'b') || (word[index] == 'c')||
(word[index] == 'd')|| (word[index] == 'e')|| (word[index] == 'f')) {
// Do nothing here
}
else {
return false;
}
index++;
}
return true;
}
Obviously now that the affirmative block of the if is empty, you could refactor a little and only check for the negative condition. The logic of it should read closely to the way you described the problem in words:
"The minute the program finds a letter that is not one of those the program should return false."
bool is_favorite(string word){
int length = word.length(); // "word" is the string.
int index = 0;
while (index < length) {
if (!is_letter_a_through_f((word[index])
return false;
index++;
}
return true;
}
I replaced your large logical check against many characters with a function in the above code to make it more readable. I trust you do that without difficulty. My own preference is to keep statements short so that they are readable, and so that when you read the code, you can hold in your short-term memory the logic of what you are saying about control flow without being overloaded by the mechanics of your letter comparison.
I am trying to print a list of invalid emailaddress (which has a space and does not have a # or .) from a list of email addresses. The list has a few email addresses which have spaces, and no '#' or '.' but still it does not print anything.
//Declaring boolean variables
bool atPresent;
bool periodPresent;
bool spacePresent;
string emailid = someemailfrom a list;
atPresent = false;
periodPresent = false;
spacePresent = false;
//looking for #
size_t foundAt = emailid.find('#');
if (foundAt != string::npos) {
atPresent = true;
}
//looking for '.'
size_t foundPeriod = emailid.find('.');
if (foundPeriod != string::npos) {
periodPresent = true;
}
//looking for ' '
size_t foundSpace = emailid.find(' ');
if (foundSpace != string::npos) {
spacePresent = true;
}
//checking to see if all conditions match
if ( (atPresent == false) && (periodPresent == false) && (spacePresent == true)) {
cout << emailid << endl;
}
(atPresent == false) && (periodPresent == false) && (spacePresent == true)
Is wrong. It is only true, when all of the three criteria for an invalid adress are met. But an address is invalid as soon as at least on criteria is met. This would be
(atPresent == false) || (periodPresent == false) || (spacePresent == true)
And simplified:
!atPresent || !periodPresent || spacePresent
replace && statements by || statements : you are only printing those which doesn't have # AND have a space AND have a period. You should use a regex, so you can do it on one line, and know how to use them is always usefull when you try to validate user data
My current assignment requires me to allow for a user to pass strings of characters and convert from roman numerals to a decimal value. I've faced an issue that when random words are passed through the program it must return the longest valid prefix and ignore the remaining characters. My code runs through a while loop and once it reaches the break condition it repeats the code again for the remaining characters. I'm unsure what next step I should take to fix this code to output correctly. Following image shows the characters to pass through the code.
Image of test failure
#include <iostream>
#include <ctype.h>
using namespace std;
int
main (){
while (!cin.eof ()){
char inputChar;
int result = 0;
while (cin.get (inputChar))
{
inputChar = toupper (inputChar);
if (inputChar == 'M')
result = result + 1000;
else if (inputChar == 'D')
{
inputChar = cin.peek ();
inputChar = toupper (inputChar);
if (inputChar == 'M')
{
result = result - 500;
continue;
}
else if (inputChar == 'D'){
result = result;
continue;
}
else
{
result = result + 500;
continue;
}
}
else if (inputChar == 'C')
{
inputChar = cin.peek ();
inputChar = toupper (inputChar);
if (inputChar == 'M' || inputChar == 'D')
{
result = result - 100;
continue;
}
else
{
result = result + 100;
continue;
}
}
else if (inputChar == 'L')
{
inputChar = cin.peek ();
inputChar = toupper (inputChar);
if (inputChar == 'M' || inputChar == 'D' || inputChar == 'C')
{
result = result - 50;
continue;
}
else
{
result = result + 50;
continue;
}
}
else if (inputChar == 'X')
{
inputChar = cin.peek ();
inputChar = toupper (inputChar);
if (inputChar == 'M' || inputChar == 'D' || inputChar == 'C'
|| inputChar == 'L')
{
result = result - 10;
continue;
}
else
{
result = result + 10;
continue;
}
}
else if (inputChar == 'V')
{
inputChar = cin.peek ();
inputChar = toupper (inputChar);
if (inputChar == 'M' || inputChar == 'D' || inputChar == 'C'
|| inputChar == 'L' || inputChar == 'X')
{
result = result - 5;
continue;
}
else
{
result = result + 5;
continue;
}
}
else if (inputChar == 'I')
{
inputChar = cin.peek ();
inputChar = toupper (inputChar);
if (inputChar == 'M' || inputChar == 'D' || inputChar == 'C'
|| inputChar == 'L' || inputChar == 'X' || inputChar == 'V')
{
result = result - 1;
continue;
}
else
{
result = result + 1;
continue;
}
}
else{
break;
}
}
if (result != 0)
{
cout << result << endl;
}
}
return 0;
}
Your break statement does not go out of the both while loops. It only breaks the inner loop, not the outer one. You can use a flag to tell the outer loop not to continue:
int main()
{
bool exit_loop = false; // define a flag here
while (!cin.eof()) {
char inputChar;
int result = 0;
if (exit_loop) // check the flag here
break;
while (cin.get(inputChar))
{
and while breaking set the flat to true:
}
else
{
exit_loop = true; // set the flag here
break;
}
Maybe try clearing the buffer at the end of the outer loop. This way the longest prefix is printed and the same word isn't being used again through the iteration.
if (result != 0)
{
cout << result << endl;
}
cin. ignore(10000, '\n'); //used to ignore or clear characters from the input buffer
}
I'm writing an NMEAParser library. As its name suggests, it parses NMEA sentences. Nothing crazy.
Its entry point is a function that accepts an NMEA string as its only parameter and looks at its beginning to pass it to the right decoder. Here is the function:
bool NMEAParser::dispatch(const char *str) {
if (!str[0]) {
return false;
}
//check NMEA string type
if (str[0] == '$') {
//PLSR245X
if (str[1] == 'P' && str[2] == 'L' && str[3] == 'S' && str[4] == 'R' && str[5] == ',' && str[6] == '2' && str[7] == '4' && str[8] == '5' && str[9] == ',') {
if (str[10] == '1')
return parsePLSR2451(str);
if (str[10] == '2')
return parsePLSR2452(str);
if (str[10] == '7')
return parsePLSR2457(str);
} else if (str[1] == 'G' && str[2] == 'P') {
//GPGGA
if (str[3] == 'G' && str[4] == 'G' && str[5] == 'A')
return parseGPGGA(str);
//GPGSA
else if (str[3] == 'G' && str[4] == 'S' && str[5] == 'A')
return parseGPGSA(str);
//GPGSV
else if (str[3] == 'G' && str[4] == 'S' && str[5] == 'V')
return parseGPGSV(str);
//GPRMC
else if (str[3] == 'R' && str[4] == 'M' && str[5] == 'C')
return parseGPRMC(str);
//GPVTG
else if (str[3] == 'V' && str[4] == 'T' && str[5] == 'G')
return parseGPVTG(str);
//GPTXT
else if (str[3] == 'T' && str[4] == 'X' && str[5] == 'T')
return parseGPTXT(str);
//GPGLL
else if (str[3] == 'G' && str[4] == 'L' && str[5] == 'L')
return parseGPGLL(str);
}
//HCHDG
else if (str[1] == 'H' && str[2] == 'C' && str[3] == 'H' && str[4] == 'D' && str[5] == 'G')
return parseHCHDG(str);
}
return false;
}
The problem I have is that this function's cyclomatic complexity is quite high, and my SonarQube complains about it:
It's not really a problem as the code is quite easy to read. But I was wondering how I could reduce its complexity while still keeping it simple to read and efficient.
You can simplify this quite a lot:
if (std::string_view{str, 10} == "$PLSR,245,")
{
switch (str[10])
{
case '1' : return parsePLSR2451(str);
case '2' : return parsePLSR2452(str);
case '7' : return parsePLSR2457(str);
}
}
else if (std::string_view{str + 1, 2} == "GP")
{
auto s = std::string_view{str + 3, 3};
if (s == "GGA")
return parseGPGGA(str);
if (s == "GSA")
return parseGPGSA(str);
// ... etc
}
else if (std::string_view{str + 1, 5} == "HCHDG")
{
return parseHCHDG(str);
}
return false;
There's no extra strings being constructed either, so it should be at least as efficient.