C++ newbie here. Writing a simple program. Everything works,except when I attempt to extract firstname and surname and print these individually, the number of letters printed in surname will always be the same size as the number in firstname. So if the name is will jackson, the program will print firstname as: will and second name as: jack. How do I solve this?
here is the code
for( i = 0; i < 19; i++)
if(cAddress[i] == ' ' || cAddress[i] == '_' || cAddress[i] == '-' || cAddress[i] == '.')
break;
strncpy(cFirst, cAddress, i);
cFirst[i] = '\0';
cout << endl << "\n";
cout << "Your first name is " << cFirst << endl;
strcpy(cSur,cAddress + i + 1);
cSur[i] = '\0';
cout << endl;
cout << "Your surname is " << cSur << endl;
You are adding a \0 character at the (i+1)th position - this marks the end of string, so that's all it prints. Before that line, cSur probably contains the entire surname.
Most of your code looks a lot like C -- and doesn't even take full advantage of the C standard library. In C++, I'd write something more like:
int pos = Address.find_first_of(" _-.");
std::string FirstName(Address, 0, pos);
std::string SurName(Address, pos);
If, for whatever reason, you insist on using C style strings, you might consider starting with strpbrk to find the separator.
cSur[i] = '\0';
is incorrect. i is the length of the first name. cSur is already zero terminated by strcpy.
Having explained that, Jerry Coffin's answer is what I would recommend as good c++ code.
Related
I'm a beginner at c++(took a couple classes, then no c++ for a while, then starting back up several months later), and I'm trying to count the number of words in a simple sentence and then count the number of numbers in that same sentence. To count the words, I use:
int countWord(char *word)
{
int counter = 0;
char *words = strtok(word, " 0123456789-");
while (words != NULL)
{
counter++;
words = strtok(NULL, " 0123456789-");
}
return counter;
}
The number counter is basically the same, just instead of using integers I use the alphabet.
char *num = strtok(number, " abcdefghijklmnopqrstuvwxyz");
My main is:
int main()
{
char str[] = "what time is 88 it 99today";
cout << "words = " << countWord(str) << " " << "numbers = " <<
countNum(str) << endl;
system("pause");
return 0;
}
When I run this it outputs: words = 3 numbers = 2.
When i rearrange main to:
char str[] = "what time is 88 it 99today";
cout << "words = " << countWord(str) << " ";
cout << "numbers = " << countNum(str) << endl;
output is: words = 5 numbers = 0
Can anyone explain why this is incorrect? Also, if anyone can refer me to a text that covers this, I'd appreciate that. The text I learned from is: "C++ Programming: Program Design Including Data Structures by D.S. Malik. I didn't see any techniques in this book to count "words". Thank you.
The issue is that strtok marks the end of tokens in the original string by a null character. Citing from cppreference:
If such character was found, it is replaced by the null character '\0' and the pointer to the following character is stored in a static location for subsequent invocations.
Notes: This function is destructive: it writes the '\0' characters in the elements of the string str. In particular, a string literal cannot be used as the first argument of strtok.
In your case the line
cout << "words = " << countWord(str) << " " << "numbers = " <<
countNum(str) << endl;
is a composition of operator<<, like
...operator<<(operator<<(cout, "words"), countWord(str))...
so the line countNum(str) is evaluated first. Then countWord(str) is evaluated secondly. This is in contrast to
cout << "words = " << countWord(str) << " ";
cout << "numbers = " << countNum(str) << endl;
where the other way around happens.
One solution is to use a copy of the original string when using strtok, e.g. use strtok(strdup(str)) every time. Better yet, use standard C++ library features, like std::string, std::count_if etc. I'm sure there are plenty of word counting solutions around using pure C++.
Vlad has submitted a nice answer for your C-style code. My answer is demonstrating use of more C++ libraries to help move things along:
#include <iostream>
#include <string>
#include <vector>
#include <regex>
int main() {
// The main string.
std::string str = "what time is 88 it 99today";
// Space is your delimiter
std::string delimiter = " ";
// Create a regex string for matching numbers, including floating point.
std::regex number_chars(std::string("[0123456789.]+"));
// A lambda function to help tokenize strings.
// Returns a vector of substring tokens.
// The internal code is taken from an answer on Stack Overflow.
auto tokenizer = [](std::string s, std::string delimiter) {
size_t pos = 0;
std::string token;
std::vector<std::string> tokens;
while (pos = (s.find(delimiter))) {
token = s.substr(0, pos);
tokens.push_back(token);
s.erase(0, pos + delimiter.length());
if (pos == std::string::npos)
break;
}
return tokens;
};
// Apply the lambda.
auto tokens = tokenizer(str, delimiter);
// Output your tokens.
for (auto it : tokens) {
std::cout << it << "\n";
} std::cout << "\n";
// Output tokens that are numbers.
for (auto it : tokens) {
if (std::regex_match(it, number_chars)) {
std::cout << "String: " << it << " is a number.\n";
}
}
return 0;
}
Since C++ has a regular expression library in C++11, it would be good to leverage it.
Coliru: http://coliru.stacked-crooked.com/a/43cd6711e1243f4a
I have the following code to open a file and read the data from it, then take the relavent part and print it to screen.
char* search = "model name";
int Offset;
string Cpu;
ifstream CpuInfo;
CpuInfo.open ("/proc/cpuinfo");
if(CpuInfo.is_open())
{
while(!CpuInfo.eof())
{
getline(CpuInfo,Cpu);
if ((Offset = Cpu.find(search, 0)) != string::npos)
{
//cout << "found '" << search << " " << line << endl;
break;
}
}
CpuInfo.close();
}
Cpu.replace (0,13,"");
cout << Cpu
This usually outputs the type of CPU your using, but one problem is that some people have various spaces inbetween the words that it prints out.
My question is how to remove all the spaces from inbetween the words. They can of random ammount and aren't always present.
Thank you in advance.
Since your question states: "how to remove all the spaces from inbetween the words":
You can use std::remove_if from the standard <algorithm> library in addition to std::isspace:
std::string mystring = "Text with some spaces";
std::remove_if(mystring.begin(), mystring.end(), std::isspace);
This now becomes:
Textwithsomespaces
REFERENCES:
http://en.cppreference.com/w/cpp/algorithm/remove
Why I cannot initialize string with "\x"
string s = "\x"
It would be useful if I could later write:
int grade = 0;
while (cin >> grade)
if (grade < 60)
cout << "Your grade letter is F!";
else {
x = 50 - grade/10;
s = s + static_cast<string>(x);
cout << "Your grade letter is " << s << endl;
}
I prefer answer, that uses escape sequences computation for setting grade letter.
Because the syntax forbids it. The \x sequence in a string literal is a prefix that means "here comes the hexadecimal code for a character", but you're trying to omit the code part. That means it's not possible to parse the literal and figure out which character to put in the string.
Note that this is a compile-time thing, it has to be possible to compute the sequence of characters represented by a string literal, by just looking at the literal itself.
You misunderstand the way the escape sequences are processed. They are computed at compile-time, not at run time. On other words, when you write
"\x48"
it does not become a string of four characters at runtime; compiler converts it to a single-character string before the program is run.
You also misunderstand static_cast<...>: if x is not a std::string, static-casting it to string will result in an error; if it is a std::string, static-casting will have no effect.
You can create a single-character string at runtime and put a character code into its only character like this:
int grade = 83; // <<<=== 1..100
grade--;
int gradeLetter = g < 60 ? 'F' : ('A' + (100-grade)/10);
// At this point you can do the output:
cout << "Your grade is " << gradeLetter << endl;
// If you must have a string, do this:
string gradeStr(1,gradeLetter);
cout << "Your grade is " << gradeStr << endl;
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.
Closed 10 years ago.
With the assistance of others, I have redone the code from scratch due to them pointing out numerous errors and things that wouldn't work. Thus I have changed the code massively.
I have the program working other than two formatting settings that I can't figure out how to get to work.
I need to only print "DAILY SCOOP REPORT" once at the top of the output, but I've moved it around but due to the way the arrays are set up I don't know where to put it.
Here is my code:
#include <iostream>
include
include
include
include
include
using namespace std;
int main()
{
string flavor_input, Capitalize;
string flavors[] = { "Chocolate", "Vanilla", "Strawberry", "Mint", "Rocky Road", "Mocha" };
int scoop_count [6] = { 0, 0, 0, 0, 0, 0 }, scoops = 0, j, k;
bool valid_option;
cout << "Welcome to Frozen Tongue Ice Cream Shop\n"<<endl;
cout << "Flavors avaliable: "<<endl;
cout << "Chocolate "<<endl;
cout << "Valnilla "<<endl;
cout << "Strawberry "<<endl;
cout << "Mint "<<endl;
cout << "Rocky Road "<<endl;
cout << "Mocha \n"<<endl;
while(true) {
cout << "Please enter the flavor of icecream sold or 'STOP' to exit.\n"<<endl;
getline (cin, flavor_input); // getline causes rocky road to be accepted with a space between the words.
string::iterator it( flavor_input.begin()); //converting the first letter of input to uppercase if not already.
if (it != flavor_input.end())
flavor_input[0] = toupper((unsigned char)flavor_input[0]);
while(++it != flavor_input.end())
{
*it = tolower((unsigned char)*it);
}
if (flavor_input == "STOP" || flavor_input == "Stop")
break;
valid_option = false;
for(int i=0;i<6;i++) //Checks to see if input matches those of possible inputs.
if(!flavor_input.compare(flavors[i]))
{
valid_option = true;
break;
}
if(!valid_option)
{
cout << "Invalid Flavor. Try again.\n\n";
flavor_input.clear();
continue;
}
for(int i=0;i<6;i++)
{
if(!flavor_input.compare(flavors[i]))
{
cout << "Enter how many scoops were sold: ";
cin >> scoops;
cin.ignore();
scoop_count[i] += scoops;
scoops = 0;
cout << '\n';
break;
}
}
}
for(int i=0;i<6;i++)
{
if(!scoop_count[i])
continue;
else
{
cout << "\nDAILY SCOOP REPORT: "<<endl;
cout << setiosflags( ios::left )
<< setw( 11 ) << flavors[i]
<< resetiosflags( ios::left )
<< setw( 4 ) << scoop_count[i] << endl;
}
}
cin.get();
return 0;
}
Thanks again for all of the assistance. It is greatly appreciated.
Thanks to all the assistance and pointing me in the direction of what to study, I have the program completed other than one last part.
I figured out that why it wasn't working when I moved the "DAILY SCOOP REPORT" line around. I had renamed the file and when I compiled it, it was outputing the "last working configuration" kinda deal if that makes sense. So I created a new project (the .cpp file has to have a certain name for submission) and put the code in it. Now the line is printed only once.
In the code block below, I have it where it lowers casing for all other letters other than the first or so it seems to be doing. The reason I have the case coding the way I do is that the instructions want the flavor report to print out with first letter of each word cap and lower after that. I am going to look into how to cap the 2nd "R" in Rocky Road, but other than the ignore white-space I don't really know how. Do I need to parse the line?
Anyone to point me in the right direction would be appreciated.
I tried but it gives error that in the first if statement "syntax error : identifier 'flavor_input'".
//converting the first letter of input to uppercase if not already.
string::iterator it( flavor_input.begin());
if flavor_input = "rocky road"
(it != flavor_input.end())
flavor_input[6] = toupper((unsigned char)flavor_input[6]);
if (it != flavor_input.end())
flavor_input[0] = toupper((unsigned char)flavor_input[0]);
while(++it != flavor_input.end())
{
*it = tolower((unsigned char)*it);
}
switch doesn't work with strings.
You need to use the operator == to select the right choice like so:
string count = // ... something
if (count == "choc") {
}
else if (count == "van") {
}
else if (count == "str") {
} // .. etc
A few other things: make sure you spell string with a consistent case, all lower case and no upper case. String is something different than string.
Make sure you surround strings with double quotes "" and not single quotes ''. single quotes are for single characters like 'a' or 'b'. double quotes are for multiple characters strings like "abc" and "hello"
Having the word count as both the function name and an argument name is probably a bad idea and will not compile because the same name means two different things.
You can't return multiple values from a function. writing something like return a,b,c; doesn't mean what you probably want it to mean. the comma (,) operator allows several expressions to be evaluated in the same statement and the result is the value of the last expression so writing return 1,2,3; is exactly the same as writing return 3;
C++ cannot switch on a string. Replace your switch(count) {...} with if/else if statements. Additionally the proper format for a string is "string", not 'string' (single quotes designate a single character, like: 'a'). Also, ensure that you always use the correct casing for string objects (string as opposed to String, like you have as your return values)
Other than that, it would be helpful to see the compiler errors you are getting.
Another thing I noticed in your source: you will have to omit the semicolons (-cola?) at the end of the following lines:
cout << "Please enter the flavor of icecream sold or 'STOP' to exit.\n"<<endl
<< "Flavors avaliable:\n"<<endl
<< "chocolate\n"<<endl
<< "valnilla\n"<<endl
<< "strawberry\n"<<endl
<< "mint\n"<<endl
<< "rocky road\n"<<endl
<< "mocha\n"<<endl
<< "Enter flavor : ";
This is just a single huge expression. The same applies to
cout << "\nDAILY SCOOP REPORT: \n"<<endl
<< "chocolate :"<<chocolate<<"\n"<<endl
<< "vanilla :"<<vanilla<<"\n"<<endl
<< "strawberry :"<<strawberry<<"\n"<<endl
<< "mint :"<<mint<<"\n"<<endl
<< "rocky road :"<<rocky_road<<"\n"<<endl
<< "mocha :"<<mocha<<"\n"<<endl;
Also: the endl and the "\n" are redundant. You will see the choices being separated by empty lines.
I haven't looked at the whole thing, but this isn't going to do what you want:
if (flavor = 'STOP' || 'stop')
I think you need:
if (flavor.compare("STOP") == 0 || flavor.compare("stop") == 0)
Let's go down the problems I see.
String count (string& flavor, string& count, string& chocolate, string& vanilla, string& strawberry, string& mint, string& rocky_road, string& mocha);
You're using String here, I'm sure you meant std::string or just string.
Inside the count function (SO is truncating the code when pasted), you're ending the line with a semicolon after endl yet trying to continue the stream output. I think you meant
Next:
if (flavor = 'STOP' || 'stop')
I think you meant to use the operator== instead of operator=, which is assignment not comparison. Also, there are no junctions in c++, so you will have to write that out as:
if (flavor == 'STOP' || flavor == 'stop')
Next:
switch (count) { case 'choc' :
Two problems here. First, you can only use plain-old-data (pod) in switch statements. Using std::string in a switch will not automatically call operator==; you will have to use if/else statements. Also, string literals are double quoted whereas character literals are single quoted.
Next:
chocCount + count;
This isn't really a statement. I'm sure you meant chocCount += count;
Next:
if (flavor = chocolate) chocCount + count;
Again, you want to use == and chocCount += count;.
Most of these problems are repeated. You should fix each of these problems everywhere they exist. There may be other problems, but I was basically compiling that in my head.
I didn't read through all of it to find semantic problems, but your count function is clearly not returning a count (at least what I currently see posted). You are returning a String, which I assume you meant string.
That's all this human compiler is going to solve for 1 homework assignment. I could recommend you go read a good C++ tutorial.
I was having some trouble with my parsing function so I put some cout statements to tell me the value of certain variables during runtime, and I believe that atoi is incorrectly converting characters.
heres a short snippet of my code thats acting strangely:
c = data_file.get();
if (data_index == 50)
cout << "50 digit 0 = '" << c << "' number = " << atoi(&c) << endl;
the output for this statement is:
50 digit 0 = '5' number = 52
I'm calling this code within a loop, and whats strange is that it correctly converts the first 47 characters, then on the 48th character it adds a 0 after the integer, on the 49th character it adds a 1, on the 50th (Seen here) it adds a two, all the way up to the 57th character where it adds a 9, then it continues to convert correctly all the way down to the 239th character.
Is this strange or what?
Just to clarify a little more i'll post the whole function. This function gets passed a pointer to an empty double array (ping_data):
int parse_ping_data(double* ping_data)
{
ifstream data_file(DATA_FILE);
int pulled_digits [4];
int add_data;
int loop_count;
int data_index = 0;
for (char c = data_file.get(); !data_file.eof(); c = data_file.get())
{
if (c == 't' && data_file.get() == 'i' && data_file.get() == 'm' && data_file.get() == 'e' && data_file.get() == '=')
{
loop_count = 0;
c = data_file.get();
if (data_index == 50)
cout << "50 digit 0 = '" << c << "' number = " << atoi(&c) << endl;
pulled_digits[loop_count] = atoi(&c);
while ((c = data_file.get()) != 'm')
{
loop_count++;
if (data_index == 50)
cout << "50 digit " << loop_count << " = '" << c << "' number = " << atoi(&c) << endl;
pulled_digits[loop_count] = atoi(&c);
}
add_data = 0;
for (int i = 0; i <= loop_count; i++)
add_data += pulled_digits[loop_count - i] * (int)pow(10.0,i);
if (data_index == 50)
cout << "50 index = " << add_data << endl;
ping_data[data_index] = add_data;
data_index++;
if (data_index >= MAX_PING_DATA)
{
cout << "Error parsing data. Exceeded maximum allocated memory for ping data." << endl;
return MAX_PING_DATA;
}
}
}
data_file.close();
return data_index;
}
atoi takes a string, i.e. a null terminated array of chars, not a pointer to a single char so this is incorrect and will get you unpredictable results.
char c;
//...
/* ... */ atoi(&c) /* ... */
Also, atoi doesn't provide any way to detect errors, so prefer strtol and similar functions.
E.g.
char *endptr;
char c[2] = {0}; // initalize c to all zero
c[0] = data.file.get(); // c[1] is the null terminator
long l = strtol(c, &endptr, 10);
if (endptr == c)
// an error occured
atoi expects a null-terminated string as an input. What you are supplying is not a null-terminated string.
Having said that, it is always worth adding that it is very difficult (if at all possible) to use atoi properly. atoi is a function that offers no error control and no overflow control. The only proper way to perform string-representation-to-number conversion in C standard library is functions from strto... group.
Actually, if you need to convert just a single character digit, using atoi or any other string conversion function is a weird overkill. As it has already been suggested, all you need is to subtract the value of 0 from your character digit value to get the corresponding numerical value. The language specification guarantees that this is a portable solution.
Nevermind, it was simply that I needed to convert the character into a string terminated by \0. I changed it to this code:
char buffer [2];
buffer[1] = '\0';
buffer[0] = data_file.get();
if (data_index == 50)
cout << "50 digit 0 = '" << buffer[0] << "' number = " << atoi(buffer) << endl;
and it worked.