So I have managed to convert the string into an int. However, in code #1, when I try to assign it to the first slot in the array and print it out, it prints '<'. Why is it doing this? I know it has something to do with ascii characters. Code #2 prints out the int 60 which is what I want.
atoi(menuAttributes[c].c_str()) = 20;
quantity[d] = 3;
string price[14];
#1
price[0] = atoi(menuAttributes[c].c_str()) * quantity[d];
cout << price[0] << endl;
#2
cout << atoi(menuAttributes[c].c_str()) * quantity[d] << endl;
Pretty much, I want price[0] to equal the int 60, not the char '<'. Thanks!
EDIT: Solved, thanks for everyones help. Noob here, apologies!
it's because of your price definition:
string price[14];
- you have defined it as an array of 14 strings, and attempting to assign to the first string in the array (price[0]) a numerical value (which is bogus from string's point of view).
Once you define your price as int price[14], then you'll get what you expect
price needs to be defined as an integer array. When you try to store the integer value 60 in price, it's implicitly converted to (char)60, which is < on the ASCII table.
Related
PROBLEM SOLVED: thanks everyone!
I am almost entirely new to C++ so I apologise in advance if the question seems trivial.
I am trying to convert a string of letters to a set of 2 digit numbers where a = 10, b = 11, ..., Y = 34, Z = 35 so that (for example) "abc def" goes to "101112131415". How would I go about doing this? Any help would really be appreciated. Also, I don't mind whether capitalization results in the same number or a different number. Thank you very much in advance. I probably won't need it for a few days but if anyone is feeling particularly nice how would I go about reversing this process? i.e. "101112131415" --> "abcdef" Thanks.
EDIT: This isn't homework, I'm entirely self taught. I have completed this project before in a different language and decided to try C++ to compare the differences and try to learn C++ in the process :)
EDIT: I have roughly what I want, I just need a little bit of help converting this so that it applies to strings, thanks guys.
#include <iostream>
#include <sstream>
#include <string>
int returnVal (char x)
{
return (int) x - 87;
}
int main()
{
char x = 'g';
std::cout << returnVal(x);
}
A portable method is to use a table lookup:
const unsigned int letter_to_value[] =
{10, 11, 12, /*...*/, 35};
// ...
letter = toupper(letter);
const unsigned int index = letter - 'A';
value = letter_to_value[index];
cout << index;
Each character has it's ASCII values. Try converting your characters into ASCII and then manipulate the difference.
Example:
int x = 'a';
cout << x;
will print 97; and
int x = 'a';
cout << x - 87;
will print 10.
Hence, you could write a function like this:
int returnVal(char x)
{
return (int)x - 87;
}
to get the required output.
And your main program could look like:
int main()
{
string s = "abcdef"
for (unsigned int i = 0; i < s.length(); i++)
{
cout << returnVal(s[i]);
}
return 0;
}
This is a simple way to do it, if not messy.
map<char, int> vals; // maps a character to an integer
int g = 1; // if a needs to be 10 then set g = 10
string alphabet = "abcdefghijklmnopqrstuvwxyz";
for(char c : alphabet) { // kooky krazy for loop
vals[c] = g;
g++;
}
What Daniel said, try it out for yourself.
As a starting point though, casting:
int i = (int)string[0] + offset;
will get you your number from character, and: stringstream will be useful too.
How would I go about doing this?
By trying to do something first, and looking for help only if you feel you cannot advance.
That being said, the most obvious solution that comes to mind is based on the fact that characters (i.e. 'a', 'G') are really numbers. Suppose you have the following:
char c = 'a';
You can get the number associated with c by doing:
int n = static_cast<int>(c);
Then, add some offset to 'n':
n += 10;
...and cast it back to a char:
c = static_cast<char>(n);
Note: The above assumes that characters are consecutive, i.e. the number corresponding to 'a' is equal to the one corresponding to 'z' minus the amount of letters between the two. This usually holds, though.
This can work
int Number = 123; // number to be converted to a string
string Result; // string which will contain the result
ostringstream convert; // stream used for the conversion
convert << Number; // insert the textual representation of 'Number' in the characters in the stream
Result = convert.str(); // set 'Result' to the contents of the stream
you should add this headers
#include <sstream>
#include <string>
Many answers will tell you that characters are encoded in ASCII and that you can convert a letter to an index by subtracting 'a'.
This is not proper C++. It is acceptable when your program requirements include a specification that ASCII is in use. However, the C++ standard alone does not require this. There are C++ implementations with other character sets.
In the absence of knowledge that ASCII is in use, you can use translation tables:
#include <limits.h>
// Define a table to translate from characters to desired codes:
static unsigned int Translate[UCHAR_MAX] =
{
['a'] = 10,
['b'] = 11,
…
};
Then you may translate characters to numbers by looking them up in the table:
unsigned char x = something;
int result = Translate[x];
Once you have the translation, you could print it as two digits using printf("%02d", result);.
Translating in the other direction requires reading two characters, converting them to a number (interpreting them as decimal), and performing a similar translation. You might have a different translation table set up for this reverse translation.
Just do this !
(s[i] - 'A' + 1)
Basically we are converting a char to number by subtracting it by A and then adding 1 to match the number and letters
I'm converting my c++ program to c. It's a simple factorial function. I used an interesting method in the c++ program in order to space it less and less each time the function called itself. This is what that looked like:
cout << setw( number * 3 ) << "" << "number is: " << number << endl;
It's not translating into c as easily as I had hoped. My idea (which is probably incorrect) is
char c = "";
printf( "%*c number is: %lo\n", number * 3, c, number);
I'm getting an error that says: warning: initialization makes integer from pointer without a cast [enabled by default]
How would you achieve this same effect in c? I really liked the way I implemented it in c++ and would like to do it in a similar way.
The warning you get has nothing to do with your printing attempts. The warning comes from
char c = "";
You cannot initialize a character variable with a string literal. If you need a character, initialize it with a character constant
char c = ' ';
If you need a string, declare it as a pointer or as a character array
const char *c = "";
char c[] = "";
It really depends on what you are trying to do.
But in reality you don't need that c at all. The direct analogue of your C++ code in C would be a mere
printf("%*snumber is: %d\n", number * 3, "", number);
assuming your number is of int type.
in C, you use single quote for one character, you use double quotes for C style strings
So
char c = "" ;
is wrong.
you should use
char c = ' ';
How would you achieve this same effect in c?
Something like:
printf("%s: %d", "The number is", number * 3";
spaces can be edited into the printf statement, but if you want a char array space you can do it like:
char c[]={" "};
printf("%s:%s%d", "The number is", c, number * 3";
I am trying to iterate through a string and copy chunks of information based off of an initial key value and a key value that identifies the end of the chunk of info. However when I try to subtract my initial and final values to find the length of the chunk im looking for, I receive a seemingly arbitrary value.
So the start and end indicies are found by:
currentstringlocation = mystring.find("value_im_looking_to_start_at, 0);
endlocation = mystring.find("value_im_looking_to_stop_at", currentstringlocation);
I'm then trying to do something like:
mystring.copy(newstring,(endlocation-currentlocation), currentlocation);
This however isn't giving me the results I want. Here's an excerpt from my code and the output it yields.
stringlocation2=topoinfo.find("\n",stringlocation+11);
topoinfo.copy(address,(stringlocation2-stringlocation+11),stringlocation+11);
cout << (stringlocation2-stringlocation+11) << "\n";
cout << stringlocation2 << "\t" << stringlocation+11 << "\n";
output:
25
59 56
So clearly the chunk of info I'm trying to capture spans 3 characters, however when I subtract the two I get 25. Can someone explain to me why this happens and how I can work around it?
You are calculating the length wrong, try instead something like:
topoinfo.copy(address, stringlocation2 - (stringlocaion + 11),
stringlocation + 11);
After this, address will contain the copied string. Remember though: If address is a character array or a character pointer, then you should add the terminating '\0' character yourself!
A better solution to get a substring is to actually use the std::string::substr function:
std::string address = topoinfo.substr(stringlocation + 11,
stringlocation2 - (stringlocaion + 11));
Should be
topoinfo.copy(address,stringlocation2-(stringlocation+11),stringlocation+11);
cout << stringlocation2-(stringlocation+11) << "\n";
You got your brackets wrong.
Hey so following this Question
I've gotten stuck again, and yeah I've tried looking through the web and through my textbook. I know its probably bad posting another question so soon, but I'm truly stumped on this problem. So anyways...
The next part of the assignment asks me to find the age of the person, this age is located on the next byte after the name. Basically if the name was "Bob" it would be
[L][u][k][e][\0][\0][1][5]
where all names with even number of characters get 2 null characters to make it even and then the next two bytes store a short integer. At the moment I have tried looking at the string length and then adding more onto the length before placing that onto the offset, but it doesnt seem to work
if (name.length() % 2 != 0) {
offset += (name.length());
age = *((short*)foo+offset);
cout << age << "\n";
} else {
offset += (name.length());
age = *((short*)foo+offset);
cout << age << "\n";
}
You are missing that C and C++ multiply pointer increments by the size of the object being pointed to. So *((short*)foo+offset) actually adds offset times sizeof(short) bytes to foo.
Or maybe you understand this but don't realise that a cast has a higher precedence than an addition, (short*)foo+offset is ((short*)foo)+offset not (short*)(foo+offset).
Anyway what you want is *(short*)((char*)foo + offset). If foo is already a char* or some similar type, then you can omit the cast to char*.
Beginners question, on loop efficiency. I've started programming in C++ (my first language) and have been using 'Principles and Practice Using C++' by Bjarne Stroustrup. I've been making my way through the earlier chapters and have just been introduced to the concept of loops.
The first exercise regarding loops asks of me the following:
The character 'b' is char('a'+1), 'c' is char('a'+2), etc. Use a loop to write out
a table of characters with their corresponding integer values:
a 97, b 98, ..., z 122
Although, I used uppercase, I created the following:
int number = 64; //integer value for # sign, character before A
char letter = number;//converts integer to char value
int i = 0;
while (i<=25){
cout << ++letter << "\t" << ++number << endl;
++i;
}
Should I aim for only having 'i' be present in a loop or is it simply not possible when converting between types? I can't really think of any other way the above can be done apart from having the character value being converted to it's integer counterpart(i.e. opposite of current method) or simply not having the conversion at all and have letter store '#'.
Following on from jk you could even use the letter itself in the loop (letter <= 'z'). I'd also use a for loop but that's just me.
for( char letter = 'a'; letter <= 'z'; ++letter )
std::cout << letter << "\t" << static_cast<int>( letter ) << std::endl;
You should aim for clarity first and you try to micro-optimize instead. You could better rewrite that as a for loop:
const int offsetToA = 65;
const int numberOfCharacters = 26;
for( int i = 0; i < numberOfCharacters; ++i ) {
const int characterValue = i + offsetToA;
cout << static_cast<char>( characterValue ) << characterValue << endl;
}
and you can convert between different types - that's called casting (the static_cast construct in the code above).
That's not a bad way to do it, but you can do it with only one loop variable like this:
char letter = 65;
while(letter <= 65+25){
printf("%c\t%d\n", letter, letter);
++letter;
}
there is nothing particularly inefficient about the way you are doing it but it certainly is possible to just convert between chars and ints (a char is an integer type). this would mean you only need to store 1 counter rather than the 3 (i, letter + number) you curently have
also, for looping from a fixed start to end a 'for' loop is perhaps more idiomatic (though its possible you havent met this yet!)
If you are concerned about the efficiency of your loop, I would urge you to try this:
Get this code compiled and running under an IDE, such as Visual Studio, and set a break point at the beginning. When you get there, switch to the disassembly view (instruction view) and start hitting the F11 (single-step) key, and keep a mental count of how many times you are hitting it.
You will see that it enters the loop, compares i against 25, and then starts doing the code for the cout line. That involves incrementing letter, and then going into the << routine for cout. It does a number of things in there, possibly going deeper into subroutines, etc., and finally comes back out, returning an object. Then it pushes "\t" as an argument and passes it to that object, and goes back in and does all the stuff it did before. Then it takes number, increments it, and passes it to the cout::<< routine that accepts an integer, calls a function to convert it to a string (which involves a loop), then does all the stuff it did before to loop that string into the output buffer and return.
Tired? You're not done yet. The endl has to be output, and when that happens, not only does it put "\n" in the buffer, but it calls the system routine to flush that buffer to the file or console where you are sending the I/O. You probably can't F11 into that, but rest assured it takes lots of cycles and doesn't return until the I/O is done.
By now, your F11-count should be in the vicinity of several thousand, more or less.
Finally, you come out and get to the ++i statement, which takes 1 or 2 instructions, and jumps back to the top of the loop to start the next iteration.
NOW, are you still worried about the efficiency of the loop?
There's an easier way to make this point, and it's just as instructive. Wrap an infinite loop around your entire code so it runs forever. While it's running, hit the "pause" button in the IDE, and look at the call stack. (This is called a "stackshot".) If you do this several times you get a good idea of how it spends time. Here's an example:
NTDLL! 7c90e514()
KERNEL32! 7c81cbfe()
KERNEL32! 7c81cc75()
KERNEL32! 7c81cc89()
MSVCRTD! 1021bed3()
MSVCRTD! 1021bd59()
MSVCRTD! 10218833()
MSVCRTD! 1023a500()
std::_Fputc() line 42 + 18 bytes
std::basic_filebuf<char,std::char_traits<char> >::overflow() line 108 + 25 bytes
std::basic_streambuf<char,std::char_traits<char> >::sputc() line 85 + 94 bytes
std::ostreambuf_iterator<char,std::char_traits<char> >::operator=() line 304 + 24 bytes
std::num_put<char,std::ostreambuf_iterator<char,std::char_traits<char> > >::_Putc() line 633 + 32 bytes
std::num_put<char,std::ostreambuf_iterator<char,std::char_traits<char> > >::_Iput() line 615 + 25 bytes
std::num_put<char,std::ostreambuf_iterator<char,std::char_traits<char> > >::do_put() line 481 + 71 bytes
std::num_put<char,std::ostreambuf_iterator<char,std::char_traits<char> > >::put() line 444 + 44 bytes
std::basic_ostream<char,std::char_traits<char> >::operator<<() line 115 + 114 bytes
main() line 43 + 96 bytes
mainCRTStartup() line 338 + 17 bytes
I did this a bunch of times, and not ONCE did it stop in the code for the outer i<=25 loop. So optimizing that loop is like someone's great metaphor: "getting a haircut to lose weight".
Since no one else mentioned it: Having a fixed amount of iterations, this is also a candidate for post-condition iteration with do..while.
char letter = 'a';
do {
std::cout << letter << "\t" << static_cast<int>( letter ) << std::endl;
} while ( ++letter <= 'z' );
However, as shown in Patrick's answer the for idiom is often shorter (in number of lines in this case).
You can promote char to int...
//characters and their corresponding integer values
#include"../../std_lib_facilities.h"
int main()
{
char a = 'a';
while(a<='z'){
cout<<a<<'\t'<<a*1<<'\n'; //a*1 => char operand promoted to integer!
++a;
}
cout<<endl;
}
Incrementing three separate variables is probably a little confusing. Here's a possibility:
for (int i = 0; i != 26; ++i)
{
int chr = 'a' + i;
std::cout << static_cast<char>(chr) << ":\t" << chr << std::endl;
}
Note that using a for loop keeps all the logic of setting up, testing and incrementing the loop variable in one place.
At this point, I wouldn't worry about micro-optimizations such as an efficient way to write a small loop like this. What you have allows a for loop to do the job nicely, but if you are more comfortable with while, you should use that. But I am not sure if that is your question.
I don't think you have understood the question properly. You are writing the code, knowing that 'A' is 65. The whole point of the exercise is to print the value of 'A' to 'Z' on your system, without knowing what value they have.
Now, to get an integer value for a character c, you can do: static_cast<int>(c). I believe that is what you're asking.
I haven't written any code because it should be more fun for you to do so.
Question for the experts: In C, I know that 'a'...'z' need not have continuous values (same for 'A'...'Z'). Is the same true for C++? I would think so, but then it seems highly unlikely that Stroustrup's book assumes that.
thanks for the help.. all i wrote down was
int main()
{
char letter = 96;
int number = letter;
int i = 0;
while(i <26)
{
cout <<++letter <<":" <<++numbers <<" ";
++i;
}
works great...and pretty simple to understand now.
I've tried this and worked fine:
char a = 'a';
int i = a; //represent char a as an int
while (a <= 'z') {
cout << a << '\t' << i << '\n';
++a;
++i;
}
Programming Principles and Practice using C++ (2nd Edition) | Bjarne Stroustrup
Chapter 4 - Computation (Try this #3 - Character Loop)
The character 'b' is char('a'+1), 'c' is char('a'+2), etc. Use
a loop to write out a table of characters with their corresponding integer values:
a 97 b 98 . . . z 122
This is how I solved the problem (from 10 years ago :D)
I am a freshmen btw, so I just started reading this book now... just want to input my solution
#include <iostream>
using namespace std;
int main()
{
int i = 0;
while (i < 26) {
cout << char('a' + i) << '\t' << int(97 + i) << '\n';
++i;
}
}
I solved it by analyzing first the problem which is knowing the char value of 'a' which is 97 up to 'z'. According to this ASCII table
https://www.ascii-code.com/#:~:text=ASCII%20printable%20characters%20%28character%20code%2032-127%29%20Codes%2032-127,digits%2C%20punctuation%20marks%2C%20and%20a%20few%20miscellaneous%20symbols.
Now, we have a clearer understanding on how to solve the said problem.