I have two strings srt1 and str2 as:
std::string str1 = "20110627120000";
std::string str2 = "20110629120000";
All I need is to convert them into time format and compare which one is greater.
And I am using below code for it but I am getting segmentation fault:
tm tm1,tm2;
sscanf(str1.c_str(),"%4d%2d%2d %2d%2d%2d",&tm1.tm_year,&tm1.tm_mon,&tm1.tm_mday,&tm1.tm_hour,&tm1.tm_min,&tm1.tm_sec);
sscanf(str2.c_str(),"%4d%2d%2d %2d%2d%2d",&tm2.tm_year,&tm2.tm_mon,&tm2.tm_mday,&tm2.tm_hour,&tm2.tm_min,&tm2.tm_sec);
std::cout << "5 \n";
if ((tm1.tm_year < tm2.tm_year) && (tm1.tm_mon<tm2.tm_mon) && (tm1.tm_mday<tm2.tm_mday ))
{
std::cout << str2 <<"is greater \n";
}
Why do you convert the strings when they are perfectly comparable?
int main()
{
std::string str1 = "20110627120000";
std::string str2 = "20110629120000";
std::cout << ((str1 < str2) ? "True" : "False") << std::endl;
}
Using sscanf with the format having a space in between "%4d%2d%2d %2d%2d%2d" looks plain wrong. Not checking the result of it, is.
Dates in the format you specified can be easily converted to integers and then compared. See below:
#include <iostream>
int main()
{
unsigned long long date_a = std::stoull("20110627120000");
unsigned long long date_b = std::stoull("20110629120000");
std::cout << std::max(date_a, date_b) << std::endl;
return 0;
}
Related
I have a piece of code:
#include <bits/stdc++.h>
using namespace std;
int main() {
//ios_base::sync_with_stdio(false);
string s[5];
s[0] = "Hello";
s[1] = "12345";
cout << s[0] << " " << s[1] << "\n";
cout << s[0][0] << " " << s[1][1] << "\n";
int y = stoi(s[1]); //This does not show an error
cout <<"y is "<< y << "\n";
//int x = stoi(s[1][1]); //This shows error
//cout <<"x is "<< x << "\n";
return 0;
}
The output of this code is:
Hello 12345
H 2
y is 12345
But it shows an error when I uncomment
int x = stoi(s[1][0]);
cout <<"x is "<< x << "\n";
If in both the cases a string is being converted to int using stoi()
function then why do the later part of code gives an error?
I have tried the same using atoi(s[1][0].c_str()) but it also gives an error.
What is the alternative for this, if I want to convert the second type of elements to int?
s[1] is a std::string, so s[1][0] is a single char in that string.
Calling std::stoi() with a char as input doesn't work because it takes only a std::string as input, and std::string doesn't have a constructor that takes just a single char as input.
To do what you are attempting, you need to do this instead:
int x = stoi(string(1, s[1][0]));
Or
int x = stoi(string(&(s[1][0]), 1));
Your call to atoi() doesn't work because you are trying to call c_str() on a single char instead of the std::string it belongs to, eg:
int x = atoi(s[1].c_str());
stoi has as input a string not a char.
Try this:
string str(s[0][0]);
int y = stoi(str);
I was just reviewing my C++. I tried to do this:
#include <iostream>
using std::cout;
using std::endl;
void printStuff(int x);
int main() {
printStuff(10);
return 0;
}
void printStuff(int x) {
cout << "My favorite number is " + x << endl;
}
The problem happens in the printStuff function. When I run it, the first 10 characters from "My favorite number is ", is omitted from the output. The output is "e number is ". The number does not even show up.
The way to fix this is to do
void printStuff(int x) {
cout << "My favorite number is " << x << endl;
}
I am wondering what the computer/compiler is doing behind the scenes.
The + overloaded operator in this case is not concatenating any string since x is an integer. The output is moved by rvalue times in this case. So the first 10 characters are not printed. Check this reference.
if you will write
cout << "My favorite number is " + std::to_string(x) << endl;
it will work
It's simple pointer arithmetic. The string literal is an array or chars and will be presented as a pointer. You add 10 to the pointer telling you want to output starting from the 11th character.
There is no + operator that would convert a number into a string and concatenate it to a char array.
adding or incrementing a string doesn't increment the value it contains but it's address:
it's not problem of msvc 2015 or cout but instead it's moving in memory back/forward:
to prove to you that cout is innocent:
#include <iostream>
using std::cout;
using std::endl;
int main()
{
char* str = "My favorite number is ";
int a = 10;
for(int i(0); i < strlen(str); i++)
std::cout << str + i << std::endl;
char* ptrTxt = "Hello";
while(strlen(ptrTxt++))
std::cout << ptrTxt << std::endl;
// proving that cout is innocent:
char* str2 = str + 10; // copying from element 10 to the end of str to stre. like strncpy()
std::cout << str2 << std::endl; // cout prints what is exactly in str2
return 0;
}
I have this array : BYTE set[6] = { 0xA8,0x12,0x84,0x03,0x00,0x00, } and i need to insert this value : "" int Value = 1200; "" ....on last 4 bytes. Practically to convert from int to hex and then to write inside the array...
Is this possible ?
I already have BitConverter::GetBytes function, but that's not enough.
Thank you,
To answer original quesion: sure you can.
As soon as your sizeof(int) == 4 and sizeof(BYTE) == 1.
But I'm not sure what you mean by "converting int to hex". If you want a hex string representation, you'll be much better off just using one of standard methods of doing it.
For example, on last line I use std::hex to print numbers as hex.
Here is solution to what you've been asking for and a little more (live example: http://codepad.org/rsmzngUL):
#include <iostream>
using namespace std;
int main() {
const int value = 1200;
unsigned char set[] = { 0xA8,0x12,0x84,0x03,0x00,0x00 };
for (const unsigned char* c = set; c != set + sizeof(set); ++c) {
cout << static_cast<int>(*c) << endl;
}
cout << endl << "Putting value into array:" << endl;
*reinterpret_cast<int*>(&set[2]) = value;
for (const unsigned char* c = set; c != set + sizeof(set); ++c) {
cout << static_cast<int>(*c) << endl;
}
cout << endl << "Printing int's bytes one by one: " << endl;
for (int byteNumber = 0; byteNumber != sizeof(int); ++byteNumber) {
const unsigned char oneByte = reinterpret_cast<const unsigned char*>(&value)[byteNumber];
cout << static_cast<int>(oneByte) << endl;
}
cout << endl << "Printing value as hex: " << hex << value << std::endl;
}
UPD: From comments to your question:
1. If you need just getting separate digits out of the number in separate bytes, it's a different story.
2. Little vs Big endianness matters as well, I did not account for that in my answer.
did you mean this ?
#include <stdio.h>
#include <stdlib.h>
#define BYTE unsigned char
int main ( void )
{
BYTE set[6] = { 0xA8,0x12,0x84,0x03,0x00,0x00, } ;
sprintf ( &set[2] , "%d" , 1200 ) ;
printf ( "\n%c%c%c%c", set[2],set[3],set[4],set[5] ) ;
return 0 ;
}
output :
1200
i have an array of string.
std::string str[10] = {"one","two"}
How to find how many strings are present inside the str[] array?? Is there any standard function?
There are ten strings in there despite the fact that you have only initialised two of them:
#include <iostream>
int main (void) {
std::string str[10] = {"one","two"};
std::cout << sizeof(str)/sizeof(*str) << std::endl;
std::cout << str[0] << std::endl;
std::cout << str[1] << std::endl;
std::cout << str[2] << std::endl;
std::cout << "===" << std::endl;
return 0;
}
The output is:
10
one
two
===
If you want to count the non-empty strings:
#include <iostream>
int main (void) {
std::string str[10] = {"one","two"};
size_t count = 0;
for (size_t i = 0; i < sizeof(str)/sizeof(*str); i++)
if (str[i] != "")
count++;
std::cout << count << std::endl;
return 0;
}
This outputs 2 as expected.
If you want to count all elements sizeof technique will work as others pointed out.
If you want to count all non-empty strings, this is one possible way by using the standard count_if function.
bool IsNotEmpty( const std::string& str )
{
return !str.empty();
}
int main ()
{
std::string str[10] = {"one","two"};
int result = std::count_if(str, &str[10], IsNotEmpty);
cout << result << endl; // it will print "2"
return 0;
}
I don't know that I would use an array of std::strings. If you're already using the STL, why not consider a vector or list? At least that way you could just figure it out with std::vector::size() instead of working ugly sizeof magic. Also, that sizeof magic won't work if the array is stored on the heap rather than the stack.
Just do this:
std::vector<std::string> strings(10);
strings[0] = "one";
strings[1] = "two";
std::cout << "Length = " << strings.size() << std::endl;
You can always use countof macro to get the number of elements, but again, the memory was allocated for 10 elements and thats the count that you'll get.
The ideal way to do this is
std::string str[] = {"one","two"}
int num_of_elements = sizeof( str ) / sizeof( str[ 0 ] );
Since you know the size.
You could do a binary search for not null/empty.
str[9] is empty
str[5] is empty
str[3] is not empty
str[4] is empty
You have 4 items.
I don't really feel like implementing the code, but this would be quite quick.
Simply use this function for 1D string array:
template<typename String, uint SIZE> // String can be 'string' or 'const string'
unsigned int NoOfStrings (String (&arr)[SIZE])
{
unsigned int count = 0;
while(count < SIZE && arr[count] != "")
count ++;
return count;
}
Usage:
std::string s1 = {"abc", "def" };
int i = NoOfStrings(s1); // i = 2
I am just wondering if we can write a template meta program for this ! (since everything is known at compile time)
A simple way to do this is to use the empty() member function of std::string like this e.g.:
size_t stringArrSize(std::string *stringArray) {
size_t num = 0;
while (stringArray->empty() != true) {
++num;
stringArray++;
}
return num;
}
This question already has answers here:
How to concatenate a std::string and an int
(25 answers)
Closed 6 years ago.
int i = 4;
string text = "Player ";
cout << (text + i);
I'd like it to print Player 4.
The above is obviously wrong but it shows what I'm trying to do here. Is there an easy way to do this or do I have to start adding new includes?
With C++11, you can write:
#include <string> // to use std::string, std::to_string() and "+" operator acting on strings
int i = 4;
std::string text = "Player ";
text += std::to_string(i);
Well, if you use cout you can just write the integer directly to it, as in
std::cout << text << i;
The C++ way of converting all kinds of objects to strings is through string streams. If you don't have one handy, just create one.
#include <sstream>
std::ostringstream oss;
oss << text << i;
std::cout << oss.str();
Alternatively, you can just convert the integer and append it to the string.
oss << i;
text += oss.str();
Finally, the Boost libraries provide boost::lexical_cast, which wraps around the stringstream conversion with a syntax like the built-in type casts.
#include <boost/lexical_cast.hpp>
text += boost::lexical_cast<std::string>(i);
This also works the other way around, i.e. to parse strings.
printf("Player %d", i);
(Downvote my answer all you like; I still hate the C++ I/O operators.)
:-P
These work for general strings (in case you do not want to output to file/console, but store for later use or something).
boost.lexical_cast
MyStr += boost::lexical_cast<std::string>(MyInt);
String streams
//sstream.h
std::stringstream Stream;
Stream.str(MyStr);
Stream << MyInt;
MyStr = Stream.str();
// If you're using a stream (for example, cout), rather than std::string
someStream << MyInt;
For the record, you can also use a std::stringstream if you want to create the string before it's actually output.
cout << text << " " << i << endl;
Your example seems to indicate that you would like to display the a string followed by an integer, in which case:
string text = "Player: ";
int i = 4;
cout << text << i << endl;
would work fine.
But, if you're going to be storing the string places or passing it around, and doing this frequently, you may benefit from overloading the addition operator. I demonstrate this below:
#include <sstream>
#include <iostream>
using namespace std;
std::string operator+(std::string const &a, int b) {
std::ostringstream oss;
oss << a << b;
return oss.str();
}
int main() {
int i = 4;
string text = "Player: ";
cout << (text + i) << endl;
}
In fact, you can use templates to make this approach more powerful:
template <class T>
std::string operator+(std::string const &a, const T &b){
std::ostringstream oss;
oss << a << b;
return oss.str();
}
Now, as long as object b has a defined stream output, you can append it to your string (or, at least, a copy thereof).
Another possibility is Boost.Format:
#include <boost/format.hpp>
#include <iostream>
#include <string>
int main() {
int i = 4;
std::string text = "Player";
std::cout << boost::format("%1% %2%\n") % text % i;
}
Here a small working conversion/appending example, with some code I needed before.
#include <string>
#include <sstream>
#include <iostream>
using namespace std;
int main(){
string str;
int i = 321;
std::stringstream ss;
ss << 123;
str = "/dev/video";
cout << str << endl;
cout << str << 456 << endl;
cout << str << i << endl;
str += ss.str();
cout << str << endl;
}
the output will be:
/dev/video
/dev/video456
/dev/video321
/dev/video123
Note that in the last two lines you save the modified string before it's actually printed out, and you could use it later if needed.
For the record, you could also use Qt's QString class:
#include <QtCore/QString>
int i = 4;
QString qs = QString("Player %1").arg(i);
std::cout << qs.toLocal8bit().constData(); // prints "Player 4"
cout << text << i;
One method here is directly printing the output if its required in your problem.
cout << text << i;
Else, one of the safest method is to use
sprintf(count, "%d", i);
And then copy it to your "text" string .
for(k = 0; *(count + k); k++)
{
text += count[k];
}
Thus, you have your required output string
For more info on sprintf, follow:
http://www.cplusplus.com/reference/cstdio/sprintf
cout << text << i;
The << operator for ostream returns a reference to the ostream, so you can just keep chaining the << operations. That is, the above is basically the same as:
cout << text;
cout << i;
cout << "Player" << i ;
cout << text << " " << i << endl;
The easiest way I could figure this out is the following..
It will work as a single string and string array.
I am considering a string array, as it is complicated (little bit same will be followed with string).
I create a array of names and append some integer and char with it to show how easy it is to append some int and chars to string, hope it helps.
length is just to measure the size of array. If you are familiar with programming then size_t is a unsigned int
#include<iostream>
#include<string>
using namespace std;
int main() {
string names[] = { "amz","Waq","Mon","Sam","Has","Shak","GBy" }; //simple array
int length = sizeof(names) / sizeof(names[0]); //give you size of array
int id;
string append[7]; //as length is 7 just for sake of storing and printing output
for (size_t i = 0; i < length; i++) {
id = rand() % 20000 + 2;
append[i] = names[i] + to_string(id);
}
for (size_t i = 0; i < length; i++) {
cout << append[i] << endl;
}
}
There are a few options, and which one you want depends on the context.
The simplest way is
std::cout << text << i;
or if you want this on a single line
std::cout << text << i << endl;
If you are writing a single threaded program and if you aren't calling this code a lot (where "a lot" is thousands of times per second) then you are done.
If you are writing a multi threaded program and more than one thread is writing to cout, then this simple code can get you into trouble. Let's assume that the library that came with your compiler made cout thread safe enough than any single call to it won't be interrupted. Now let's say that one thread is using this code to write "Player 1" and another is writing "Player 2". If you are lucky you will get the following:
Player 1
Player 2
If you are unlucky you might get something like the following
Player Player 2
1
The problem is that std::cout << text << i << endl; turns into 3 function calls. The code is equivalent to the following:
std::cout << text;
std::cout << i;
std::cout << endl;
If instead you used the C-style printf, and again your compiler provided a runtime library with reasonable thread safety (each function call is atomic) then the following code would work better:
printf("Player %d\n", i);
Being able to do something in a single function call lets the io library provide synchronization under the covers, and now your whole line of text will be atomically written.
For simple programs, std::cout is great. Throw in multithreading or other complications and the less stylish printf starts to look more attractive.
You also try concatenate player's number with std::string::push_back :
Example with your code:
int i = 4;
string text = "Player ";
text.push_back(i + '0');
cout << text;
You will see in console:
Player 4
You can use the following
int i = 4;
string text = "Player ";
text+=(i+'0');
cout << (text);
If using Windows/MFC, and need the string for more than immediate output try:
int i = 4;
CString strOutput;
strOutput.Format("Player %d", i);