I'm a newbie in C++ learning the language and playing around. I wrote a piece of code which behavior I don't understand. Could someone explain why the code below prints out random junk and not the first character of the first string in the list (that is a).
#include <iostream>
#include <vector>
#include <string>
#include <cstdlib>
#include <ctime>
#include <climits>
#include <stdio.h>
char* str2char(std::string str)
{
char cset[str.size()+1]; // +1 for the null character
for(int i = 0; i < str.size(); i++)
{
cset[i] = str[i];
}
cset[str.size()] = '\0';
return cset;
}
int main (int argc, char * const argv[]) {
std::vector< std::string > ladontakadet;
ladontakadet.push_back("aabcbbca");
ladontakadet.push_back("abcdabcd");
ladontakadet.push_back("cbbdcdaa");
ladontakadet.push_back("aadcbdca");
ladontakadet.push_back("cccbaaab");
ladontakadet.push_back("dabccbaa");
ladontakadet.push_back("ccbdcbad");
ladontakadet.push_back("bdcbccad");
ladontakadet.push_back("ddcadccb");
ladontakadet.push_back("baccddaa");
std::string v = ladontakadet.at(0);
char *r;
r = str2char(v);
std::cout << r[0] << std::endl;
return 0;
}
Why is my returning garbage, when I'm expecting it to output a?
Thnx for any help!
P.S. The output of this code is random. It doesn't always print the same character..:S
It's because you return a pointer to a local variable, a local variable that goes out of scope when the function returns.
You are already using std::string for the argument, use it instead of the array and the return pointer.
If your aim is to pass the content of a std::string to a function modifying the content of a char*:
#include <iostream>
#include <vector>
void f(char* s) {
s[0] = 'H';
}
std::vector<char> to_vector(const std::string& s) {
return std::vector<char>(s.c_str(), s.c_str() + s.size() + 1);
}
int main(void)
{
std::string s = "_ello";
std::vector<char> t = to_vector(s);
f(t.data());
std::cout << t.data() << std::endl;
}
Your function is returning garbage because you're returning the address of a local variable which goes out of scope after your function returns. It should probably look like this:
char* str2char(const std::string &str)
{
char *const cset = new char[str.size() + 1]; // +1 for the null character
strcpy(cset, str.c_str());
return cset;
}
You will need to delete your variable r by doing delete[] r;. Ideally though you wouldn't be using raw pointers, and you would use std::string for everything, or wrap the char * in a std::unique_ptr.
Related
I am studying pointers in C++. I have studied call by value and call by reference concept. I am trying to create a function to reverse a string which accepts a pointer to string and the size of string. The code is as follow
void reverse(string* str, int size)
{
int start = 0;
int end = size - 1;
while(start < end)
{
swap(*str[start++], *str[end--]);
}
}
int main()
{
string str = "Something";
reverse(&str, str.length());
cout << "Reversed string: " << str << endl;
return 0;
}
I am getting this error:
error: no match for ‘operator*’ (operand type is ‘std::string’ {aka
‘std::__cxx11::basic_string’})
12 | swap(*str[start++], *str[end--]);
I don't want to use the character array, is there way to do it?
Someone please explain, what's wrong in my code. Thank you.
Here is the simple fix. You don't need to change anything except a few lines.
#include <iostream>
#include <algorithm>
#include <cstring>
void reverse( std::string* str ) // no need to pass size to this function
{
int start = 0;
int end = str->length() - 1; // get the length of str like this
char* ptrToCharArray = const_cast<char*>( str->c_str() ); // gets the pointer to str's internal buffer
while ( start < end )
{
std::swap( ptrToCharArray[start++], ptrToCharArray[end--] ); // no need to use * operator anymore
}
}
int main()
{
std::string str = "Something";
reverse( &str );
std::cout << "Reversed string: " << str << std::endl;
return 0;
}
Output is:
Reversed string: gnihtemoS
Hopefully, this helps you.
Just need a little bit of change in your code
Change this *str[start++] to (*str).at(start++)
void reverse(string* str, int size)
{
int start = 0;
int end = size - 1;
while(start < end)
{
swap((*str).at(start++),(*str).at(end--));
}
}
int main()
{
string str = "Something";
reverse(&str, str.length());
cout << "Reversed string: " << str << endl;
return 0;
}
Note that there is no need to pass the size of the string as an argument to the function. You can use the member function std::string::size for that purpose as shown below:
Version 1: Passing pointer to string as argument
#include <iostream>
#include <algorithm>
void reverse(std::string *str)
{
int n=(*str).size()-1;//dereference the pointer and use size member function on the resulting string object
for(int i=0;i<((*str).size()/2);i++){
//Using the swap method to switch values at each index
std::swap((*str).at(i),(*str).at(n)); //note this can also be written as std::swap((*str)[i],(*str)[n]);
n = n-1;
}
}
int main()
{
std::string myString = "myString";
reverse(&myString);
std::cout<<"Reversed string is: "<<myString<<std::endl;
return 0;
}
In version 1, *(str) gives us a std::string type object. Next we call size member function on this std::string object. Similarly we can call the std::string::at member function on this std::string object.
Version 2: Passing reference to string as argument
#include <iostream>
#include <algorithm>
void reverse( std::string &str)
{
int n=str.size()-1;
for(int i=0;i<(str.size()/2);i++){
//Using the swap method to switch values at each index
std::swap(str.at(i),str.at(n));
n = n-1;
}
}
int main()
{
std::string myString = "myString";
reverse(myString);
std::cout<<"Reversed string is: "<<myString<<std::endl;
return 0;
}
This is although a code specific question but the output is quite bizarre.
I am aware of STL string etc. I was fooling around when I noticed something strange, and could not find a reason for it. :(
See the Two Codes below and the output.
[Code #1] (https://ideone.com/ydB8sQ)
#include <iostream>
#include <vector>
#include <cstdlib>
#include <cstdio>
using namespace std;
class str
{
private:
vector<char> A;
public:
str(const char *S) {
int sz = sizeof(S);
cerr << sz << endl;
for (int i = 0; i < sz; ++i) {
cout << S[i];
//A.push_back(S[i]); //!-- Comment --!//
}
}
};
int main(int argc, char const *argv[])
{
str A("");
return 0;
}
In this, An Empty String is passed and is printed. The Vector A does nothing but is relevant to this problem. In the first version, A is untouched, and the code prints garbage value. (see ideone O/P)
In this second version ( see A.push_back is now uncommented )
[Code #2] (https://ideone.com/PPHGZy)
#include <iostream>
#include <vector>
#include <cstdlib>
#include <cstdio>
using namespace std;
class str
{
private:
vector<char> A;
public:
str(const char *S) {
int sz = sizeof(S);
cerr << sz << endl;
for (int i = 0; i < sz; ++i) {
cout << S[i];
A.push_back(S[i]);
}
}
};
int main(int argc, char const *argv[])
{
str A("G");
return 0;
}
The Output is :
Gvector
This is across GCC / MinGW x64. This one never prints garbage value but always contains the word 'vector'.
Where is the char* in the function pointing to?
Why would 'vector' be there anyways?
Also, the size of char * is 8.
EDIT : This does not happen if it isn't wrapped around a 'class'.
The word 'vector' appears always. I supposed it was random garbage value but then how come ideone still has the same word in its memory?
The main problem in your code is in line int sz = sizeof(S);. sizeof(S) is always equal to sizeof(char *) which seems to be 8 on your system. sizeof gives you number of bytes for variable itself. If you want to know number of bytes in string to which your char pointer points, you should use strlen function instead.
You get that vector string in output randomly, as you are accessing memory which is not in allocated space. Accessing such memory is undefined behavior, so you get your undefined result.
Would like to generate a string from a function, in order to format some data, so the function should return a string.
Tried to do the "obvious", shown below, but this prints garbage:
#include <iostream>
#include <string>
char * hello_world()
{
char res[13];
memcpy(res, "Hello world\n", 13);
return res;
}
int main(void)
{
printf(hello_world());
return 0;
}
I think this is because the memory on the stack used for the res variable, defined in the function, is overwritten before the value can be written, maybe when the printf call uses the stack.
If I move char res[13]; outside the function, thus makes it global, then it works.
So is the answer to have a global char buffer (string) that can be used for the result?
Maybe doing something like:
char * hello_world(char * res)
{
memcpy(res, "Hello world\n", 13); // 11 characters + newline + 0 for string termination
return res;
}
char res[13];
int main(void)
{
printf(hello_world(res));
return 0;
}
Don't bother with that early-20th century stuff. By the end of the previous century we already had std::string, and that's straightforward:
#include <iostream>
#include <string>
std::string hello_world()
{
return "Hello world\n";
}
int main()
{
std::cout << hello_world();
}
You are programming c. That's not bad, but your question is about c++ so this is the solution for the question you asked:
std::string hello_world()
{
std::string temp;
// todo: do whatever string operations you want here
temp = "Hello World";
return temp;
}
int main()
{
std::string result = hello_world();
std::cout << result << std::endl;
return 0;
}
Best solution would be to use std::string. However, if you must use an array, then it is best to allocate it in the calling function (in this case, main()):
#include <iostream>
#include <cstring>
void hello_world(char * s)
{
memcpy(s, "Hello world\n", 13);
}
int main(void)
{
char mys[13];
hello_world(mys);
std::cout<<mys;
return 0;
}
Still, if you want to write a pure C code, will can do something like that.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *HelloWorld(char *s, int size)
{
sprintf(s, "Hello world!\n");
return s;
}
int main (int argc, char *argv[])
{
char s[100];
printf(HelloWorld(s, 100));
return 0;
}
Let's say you have:
const char * something = "m";
How would one make this uppercase, using toupper (or something else, if applicable)?
I want to use a char * instead of a string (I can use a string, but then I have to use str.c_str()).
So, how can I make char * something = "m"; contain "M"?
I find you choice of C strings disturbing.. but anyway.
You can't change a string literal (char *something). Try an array:
char something[] = "m";
something[0] = toupper(something[0]);
To change an entire string:
char something[] = "hello";
char *p = something;
while (*p) {
*p = toupper(*p);
p++;
}
As explained in the very famous C book - The C Programming Language by Kernighan & Ritchie in section 5.5 Character Pointers and Functions,
char amessage[] = "now is the time"; /* an array */
char *pmessage = "now is the time"; /* a pointer */
`amessage` is an array, just big enough to hold the
sequence of characters and `'\0'` that initializes it.
Individual characters within the array may be changed
but `amessage` will always refer to the same storage.
On the other hand, `pmessage` is a pointer, initialized
to point to a string constant; the pointer may subsequently
be modified to point elsewhere, but the result is undefined
if you try to modify the string contents.
OTOH, in C, to convert to upper case letters, you can use the following program as a reference.
#include <stdio.h>
#include <ctype.h>
int main(void)
{
int i=0;
char str[]="Test String.\n";
char c;
while (str[i]) {
c=str[i];
putchar(toupper(c));
i++;
}
return 0;
}
In C++
#include <iostream>
#include <string>
#include <locale>
using namespace std;
int main ()
{
locale loc;
string str="Test String.\n";
for (size_t i=0; i<str.length(); ++i)
cout << toupper(str[i],loc);
return 0;
}
EDIT: Adding pointer version (as requested by #John) for the C version
#include <stdio.h>
#include <ctype.h>
int main(void)
{
int i=0;
char str[]="Test String.\n";
char *ptr = str;
while (*ptr) {
putchar(toupper(*ptr));
ptr++;
}
return 0;
}
Hope it helps!
You can use the same algorithmic approach that you know for std::string for raw arrays:
char s[] = "hello world";
std::transform(s, s + std::strlen(s), s, static_cast<int(*)(int)>(std::toupper));
You cannot do this for immutable string literals (like const char * s = "hello world;") for obvious reasons, so you won't get around an additional allocation/copy for that.
Update: As Ildjarn says in the comment, it's important to note that string literals are always read-only, even though for historical reasons you are allowed to bind them to a pointer-to-mutable, like char * s = "hello world";. Any decent C++ compiler should slap you in the face if you attempt this, but it is valid C++ -- but any attempt to actually modify any element of s is undefined behaviour.
You can convert C-string to std::string and then use boost::to_upper to change string in place or boost::to_upper_copy to create upper case copy of the string. Here is the code example:
#include <iostream>
#include <boost/algorithm/string/case_conv.hpp>
int main ()
{
char const * s = "Test String.\n";
std::string str(s);
std::cout << boost::to_upper_copy(str).c_str() << std::endl;
return 0;
}
Hope this helps.
You could do:
#include <algorithm>
#include <iterator>
#include <ctype.h>
char test[] = "m";
std::transform(std::begin(test), std::end(test), std::begin(test), ::topper);
This applies the ::toupper function to character of the string. This is the ::toupper function in the global namespace that comes from C. std::toupper has multiple overloads and ::toupper looks more elegant than static_cast<int (*)(int)>(&std::toupper).
I know how to program in C# and VB but not have idea about how to use C++ and have to program a little exe to a barcode scanner that use C++ :(
In this moment I try to parse a scanned barcode that have multiple data sepparated with a "/", I find that exist a strtok function, tested it "manually" and worked ok but I not implemented yet a working function to call it correctly, what I have now:
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
int elemStr(char *str, char sep)
{
int cantElem;
unsigned ich;
cantElem = 0;
if (strlen(str) > 0) //at least 1 elem
cantElem++;
for (ich = 0; ich < strlen(str); ich++)
{
if (str[ich] == sep)
cantElem++;
}
return cantElem;
}
char* getElemStr(char *str, char sep[], int elem)
{
char *tempStr = NULL;
char *tok;
int currElem = 1;
// 1st data
strcpy( tempStr, str);
tok = strtok( tempStr, sep);
while( currElem != elem )
{
// Get next tokens:
tok = strtok( NULL, sep );
currElem++;
}
return tok;
}
void main( void )
{
char barcode[] = "710015733801Z/1/35";
char sep[] = "/";
char sep1 = sep[0];
char barcd[20];
char piezaChar[4];
int pieza;
char mtsChar[4];
int cantElem;
cantElem = elemStr(barcode, sep1 );
if (cantElem >= 1)
{
strcpy(barcd, getElemStr(barcode,sep,1) ); //pasa a str resultado;
printf("Cod: %s\n", barcd ); //STACK OVERFLOW HERE!
}
}
if I use strtok witout a function "getElemStr" it work ok but I try to use it on other places too.
Can I use strtok like this? You have a working example?
pd: I not have idea about pointers (sorry), good doc to learn about that?
Since you specifically asked about C++, I'm going to ignore your very c-style code and show you how to do this in C++:
#include <boost/algorithm/string.hpp>
#include <iostream>
#include <string>
#include <vector>
int main()
{
std::string barcode = "710015733801Z/1/35";
std::string sep = "/";
std::vector<std::string> v;
boost::split(v, barcode, boost::is_any_of(sep));
for(size_t i=0; i<v.size(); ++i)
std::cout << v[i] << '\n';
}
strtok destroys your original string. So i don't think it can be used with a char* that points to a static string. Static strings get copied to a read only portion of the executable.
Here is a C++ solution that doesn't use boost.
#include <string>
#include <sstream>
#include <iostream>
int main()
{
std::string barcode = "710015733801Z/1/35";
std::stringstream ss(barcode);
std::string elem;
while(std::getline(ss, elem, '/'))
{
//do something else meaningful with elem
std::cout << elem << std::endl;
}
return 0;
}
Output:
710015733801Z
1
35