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suppose I've this code:
string str[] = {"devil", "chandra"};
// check if str[0] has properly devil, character by character or not without extra variable
Now I want to check str[ 0 ]'s all character which is 'd','e','v','i','l' one by one without extra variable.
with extra variable code will be :
string n1 = "devil";
for(int i=0; i<1; i++){
string s1 = str[i]
for(int j=0; j<s1.size(); j++){
if(s1[i] == n[i]){
cout << s1[i] << " ";
}
}
Basically, I want O(n) loop where I can access all indexes string and among them all characters.
Like s[ i ] is "devil" and s[[i]] = 'd' something like this, Know it's not valid, but is there any way to do that??
Even I don't know is it a valid question or not!
I'm not sure why you would need an extra variable. If you need a conditional that checks that the first value in the array of strings is "devil", it shouldn't be anymore complicated than:
if (str[0] == "devil")
{
* Do things *
}
C++ can check a standard string all at once. You don't need to check each individual character if that's what you're thinking.
Keep in mind, this isn't going to account for situations where the string is not exactly the same. For instance, if str[0] has "Devil" instead of "devil", then the conditional will evaluate to false.
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I am trying to add "+", "-", "*" etc. to a string but it doesn't work.
Lets say I have string "12 3 +" Then I use string tmp to get values between spaces like "12" "3" "+". my cout prints "12" and "3", but "+" is missing
int ONP() {
string wyrazenie;
getline(cin, wyrazenie);
string tmp;
for (int i = 0; i < wyrazenie.length(); ++i) {
if (!isspace(wyrazenie[i])) {
tmp += wyrazenie[i];
} else {
cout << tmp << endl;
SOME CODE.....
tmp.clear();
}
}
}
Issue is that + is your last character, so you won't go in else block for it.
std::cout temp after the loop would show your missing character:
Demo
Your loop will never display the last token of the string unless the string ends in a space. When you have "12 3 +" you read in the 12 see a space, print the 12 and clear the string. You do the same thing for 3. Then you get + but since that is the last character in the string you never run the else part of the if statement to print it out. You can fix this a few ways. You can check if temp is not empty after the loop and if it is not then handle that. You can rework your logic in the loop to handle when you are on the last character and it is not a space. You could add a space to the end of the input so it will end with a space and the loop works as is.
From the code you provided, I see an issue that would cause the last character to not be printed. This is because you are only printing tmp when the next character is a space. So "12 3 +" would print "12", "3". Then tmp contains the value "+" since its never printed nor cleared, but is never printed. If your input string is "12 3 + " (notice the space) the '+' char would be printed too.
This can be solved with printing and clearing tmp after the loop is done if tmp still contains any data.
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I very new at this and have an assignment in which I would like for a loop to exit if the user inputs(trans) 'e' but also end if a calculation balance(bal) is less than a constant I have set. Basically as my question states one is a character and the other an integer, will that work? I'm not trying to get people to do my homework for me, so I'm not posting all of my code or assignment, hope it makes sense.
This is the line of code I have
do {
ask user input(&trans)
e or calculation
{
while (trans != 'e'| bal < -OVR);
Just use regular unconditional loop and multiple exit conditions:
while( true ) {
char trans;
std::cin >> trans;
if( !std::cin or trans == 'e' )
break;
calculation;
if( bal > -0VR )
break;
}
So first of all you would not do unnecessary calculations, but what is more important you would make your code more readable and easier to understand - you make loop exit decision where it should be instead of pushing it into the end.
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If I wanted to iterate through a string, and stop once I found either the letter "o" or the end of the string, would it make any difference to use a while loop to check the 2 conditions, or a for loop with 2 conditions?
const int STRING_SIZE = 16;
char str[STRING_SIZE] = "a bunch of words";
for loop
for (int i = 0; i < STRING_SIZE && str[i] != 'o'; i++){
//do something
}
while loop
int i = 0;
while (i < STRING_SIZE && str[i] != 'o'){
//do something
i++
}
Is there a performance difference between the two? Is one better practice than the other?
There is no difference in performance between the two loops except that:
for() Checks condition then if true its body is executed. So for is simile to while loop.
do-while loop works a slightly different: It executes then checks so at least an execution is ensured.
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Hi I used to solve string char uniqueness using map in C++. I found this solution somewhere and working fine. But I can not understand how it is working. Please some one explain.
bool isUnique(string s){
int check = 0;
for(int i=0;i<s.length();++i){
if(s[i] != ' '){
int val = s[i]-'a';
if( (check & ( 1 << val)) > 0) return false;
check = check | (1 << val);
}
}
return true;
}
It returns true if string has no repeated character excluding spaces otherwise returns false.
It is using an int as if it were a bitmap. A bitmap is certainly a better data structure for a character uniqueness test than a map. An int is a crude and questionable (in this case) substitute for a bitmap.
Assume an int has 32 bits. Those bits are allocated in this code for the first 32 characters beginning with lower case 'a'. So the upper case letters and most special characters have no bit positions and are treated as unique by this code even if they are not unique.
If you only care about uniqueness for lower case letters, and you are sure the code is only used in architectures that have at least 32 bits in an int, then this is a decent approach. Otherwise, when you want an array of bits, use some actual array of bits.
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I need to separate users input into units and store them in an array.
e.g if user enters 6547. Array will store {6,5,4,7} Using C++ on Linux
I would appreciate if you can help me with pseudocode or explain an algorithm.
I'm a beginner so please restrain from advising advanced function (and explain its use if you do) as we have studied basics so far
N.B| If such question has already been answered and I skipped it in search, please do point me to it.
The math for isolating right most digit:
digit = digit % 10;
The math for shifting a number right by one digit:
new_number = old_number / 10;
Every letter and number can be represented as a text character. For example, '5' is a character representing the single decimal digit 5.
The math for converting a textual digit (character) to numeric:
digit = char_digit - '0';
Example:
digit = '9' - '0';
The math for converting a numeric digit to a textual digit (character):
char_digit = digit + '0';
Example:
char_digit = 5 + '0';
Your problem basically breaks into few parts, which you need to figure out:
how to read one character from input
how to convert one character to the digit it represents
how to store it in the array
Please, try to explain if you have problem with some particular point from the list above or there is a problem somewhere else.
Suppose Variable input_string holds the number entered by the user & you want to store it in an array named 'a'...Here's a C snippet.. you can easily convert it into C++ code..
I would recommend taking the input as string rather than int so that you can directly insert the digits extracted from the end...(else you can start storing the integer from beginning and then reverse the array)
scanf("%s",&input_string)
size = strlen(input_string)-1
input = atoi(input_string)
while (input/10>0)
{
i=input%10;
input=input/10;
a[size]=i;
size--;
}
Hope that helps!
Here's a C++11 solution:
std::string input;
std::cin >> input;
int num = std::stoi(input);
std::vector<int> v_int;
for (unsigned int i = 0; i < input.size(); i++)
{
v_int.push_back(num % 10);
num /= 10;
}
// To get the numbers in the original order
std::sort(v_int.rbegin(), v_int.rend());
for (unsigned int i = 0; i < v_int.size(); i++) {
std::cout << v_int[i] << std::endl;
}
If you want it in a c-style array, do this:
int* dynamic_array = new int[v_int.size()];
std::copy(v_int.begin(), v_int.end(), dynamic_array);
delete dynamic_array;