Related
I'm trying to compare a character array against a string like so:
const char *var1 = " ";
var1 = getenv("myEnvVar");
if(var1 == "dev")
{
// do stuff
}
This if statement never validates as true... when I output var1 it is "dev", I was thinking maybe it has something to do with a null terminated string, but the strlen of "dev" and var1 are equal... I also thought maybe var1 == "dev" was comparing "dev" against the memory location of var1 instead of the value. *var1 == "dev" results in an error.... tried many things, probably a simple solution for the saavy c++ developer (I havent coded c++ in a looong time).
edit:
we've tried
if(strcmp(var1, "dev") == 0)
and
if(strncmp(var1, "dev", 3) == 0)
Thanks
edit: After testing at home I'm just going to suggest my co-worker changes the datatype to a string. I believe he was comparing a char array of a large size against a string. I put together a program that outputs sizeof, strlen, etc to help us work through it. Thanks to everyone for the help.
Use strcmp() to compare the contents of strings:
if (strcmp(var1, "dev") == 0) {
}
Explanation: in C, a string is a pointer to a memory location which contains bytes. Comparing a char* to a char* using the equality operator won't work as expected, because you are comparing the memory locations of the strings rather than their byte contents. A function such as strcmp() will iterate through both strings, checking their bytes to see if they are equal. strcmp() will return 0 if they are equal, and a non-zero value if they differ. For more details, see the manpage.
You're not working with strings. You're working with pointers.
var1 is a char pointer (const char*). It is not a string. If it is null-terminated, then certain C functions will treat it as a string, but it is fundamentally just a pointer.
So when you compare it to a char array, the array decays to a pointer as well, and the compiler then tries to find an operator == (const char*, const char*).
Such an operator does exist. It takes two pointers and returns true if they point to the same address. So the compiler invokes that, and your code breaks.
IF you want to do string comparisons, you have to tell the compiler that you want to deal with strings, not pointers.
The C way of doing this is to use the strcmp function:
strcmp(var1, "dev");
This will return zero if the two strings are equal. (It will return a value greater than zero if the left-hand side is lexicographically greater than the right hand side, and a value less than zero otherwise.)
So to compare for equality you need to do one of these:
if (!strcmp(var1, "dev")){...}
if (strcmp(var1, "dev") == 0) {...}
However, C++ has a very useful string class. If we use that your code becomes a fair bit simpler. Of course we could create strings from both arguments, but we only need to do it with one of them:
std::string var1 = getenv("myEnvVar");
if(var1 == "dev")
{
// do stuff
}
Now the compiler encounters a comparison between string and char pointer. It can handle that, because a char pointer can be implicitly converted to a string, yielding a string/string comparison. And those behave exactly as you'd expect.
In this code you are not comparing string values, you are comparing pointer values. If you want to compare string values you need to use a string comparison function such as strcmp.
if ( 0 == strcmp(var1, "dev")) {
..
}
"dev" is not a string it is a const char * like var1. Thus you are indeed comparing the memory adresses. Being that var1 is a char pointer, *var1 is a single char (the first character of the pointed to character sequence to be precise). You can't compare a char against a char pointer, which is why that did not work.
Being that this is tagged as c++, it would be sensible to use std::string instead of char pointers, which would make == work as expected. (You would just need to do const std::string var1 instead of const char *var1.
There is more stable function, also gets rid of string folding.
// Add to C++ source
bool string_equal (const char* arg0, const char* arg1)
{
/*
* This function wraps string comparison with string pointers
* (and also works around 'string folding', as I said).
* Converts pointers to std::string
* for make use of string equality operator (==).
* Parameters use 'const' for prevent possible object corruption.
*/
std::string var0 = (std::string) arg0;
std::string var1 = (std::string) arg1;
if (var0 == var1)
{
return true;
}
else
{
return false;
}
}
And add declaration to header
// Parameters use 'const' for prevent possible object corruption.
bool string_equal (const char* arg0, const char* arg1);
For usage, just place an 'string_equal' call as condition of if (or ternary) statement/block.
if (string_equal (var1, "dev"))
{
// It is equal, do what needed here.
}
else
{
// It is not equal, do what needed here (optional).
}
Source: sinatramultimedia/fl32 codec (it's written by myself)
your thinking about this program below
#include <stdio.h>
#include <string.h>
int main ()
{
char str[][5] = { "R2D2" , "C3PO" , "R2A6" };
int n;
puts ("Looking for R2 astromech droids...");
for (n=0 ; n<3 ; n++)
if (strncmp (str[n],"R2xx",2) == 0)
{
printf ("found %s\n",str[n]);
}
return 0;
}
//outputs:
//
//Looking for R2 astromech droids...
//found R2D2
//found R2A6
when you should be thinking about inputting something into an array & then use strcmp functions like the program above ... check out a modified program below
#include <iostream>
#include<cctype>
#include <string.h>
#include <string>
using namespace std;
int main()
{
int Students=2;
int Projects=3, Avg2=0, Sum2=0, SumT2=0, AvgT2=0, i=0, j=0;
int Grades[Students][Projects];
for(int j=0; j<=Projects-1; j++){
for(int i=0; i<=Students; i++) {
cout <<"Please give grade of student "<< j <<"in project "<< i << ":";
cin >> Grades[j][i];
}
Sum2 = Sum2 + Grades[i][j];
Avg2 = Sum2/Students;
}
SumT2 = SumT2 + Avg2;
AvgT2 = SumT2/Projects;
cout << "avg is : " << AvgT2 << " and sum : " << SumT2 << ":";
return 0;
}
change to string except it only reads 1 input and throws the rest out
maybe need two for loops and two pointers
#include <cstring>
#include <iostream>
#include <string>
#include <stdio.h>
using namespace std;
int main()
{
char name[100];
//string userInput[26];
int i=0, n=0, m=0;
cout<<"your name? ";
cin>>name;
cout<<"Hello "<<name<< endl;
char *ptr=name;
for (i = 0; i < 20; i++)
{
cout<<i<<" "<<ptr[i]<<" "<<(int)ptr[i]<<endl;
}
int length = 0;
while(name[length] != '\0')
{
length++;
}
for(n=0; n<4; n++)
{
if (strncmp(ptr, "snit", 4) == 0)
{
cout << "you found the snitch " << ptr[i];
}
}
cout<<name <<"is"<<length<<"chars long";
}
I'm trying to compare a character array against a string like so:
const char *var1 = " ";
var1 = getenv("myEnvVar");
if(var1 == "dev")
{
// do stuff
}
This if statement never validates as true... when I output var1 it is "dev", I was thinking maybe it has something to do with a null terminated string, but the strlen of "dev" and var1 are equal... I also thought maybe var1 == "dev" was comparing "dev" against the memory location of var1 instead of the value. *var1 == "dev" results in an error.... tried many things, probably a simple solution for the saavy c++ developer (I havent coded c++ in a looong time).
edit:
we've tried
if(strcmp(var1, "dev") == 0)
and
if(strncmp(var1, "dev", 3) == 0)
Thanks
edit: After testing at home I'm just going to suggest my co-worker changes the datatype to a string. I believe he was comparing a char array of a large size against a string. I put together a program that outputs sizeof, strlen, etc to help us work through it. Thanks to everyone for the help.
Use strcmp() to compare the contents of strings:
if (strcmp(var1, "dev") == 0) {
}
Explanation: in C, a string is a pointer to a memory location which contains bytes. Comparing a char* to a char* using the equality operator won't work as expected, because you are comparing the memory locations of the strings rather than their byte contents. A function such as strcmp() will iterate through both strings, checking their bytes to see if they are equal. strcmp() will return 0 if they are equal, and a non-zero value if they differ. For more details, see the manpage.
You're not working with strings. You're working with pointers.
var1 is a char pointer (const char*). It is not a string. If it is null-terminated, then certain C functions will treat it as a string, but it is fundamentally just a pointer.
So when you compare it to a char array, the array decays to a pointer as well, and the compiler then tries to find an operator == (const char*, const char*).
Such an operator does exist. It takes two pointers and returns true if they point to the same address. So the compiler invokes that, and your code breaks.
IF you want to do string comparisons, you have to tell the compiler that you want to deal with strings, not pointers.
The C way of doing this is to use the strcmp function:
strcmp(var1, "dev");
This will return zero if the two strings are equal. (It will return a value greater than zero if the left-hand side is lexicographically greater than the right hand side, and a value less than zero otherwise.)
So to compare for equality you need to do one of these:
if (!strcmp(var1, "dev")){...}
if (strcmp(var1, "dev") == 0) {...}
However, C++ has a very useful string class. If we use that your code becomes a fair bit simpler. Of course we could create strings from both arguments, but we only need to do it with one of them:
std::string var1 = getenv("myEnvVar");
if(var1 == "dev")
{
// do stuff
}
Now the compiler encounters a comparison between string and char pointer. It can handle that, because a char pointer can be implicitly converted to a string, yielding a string/string comparison. And those behave exactly as you'd expect.
In this code you are not comparing string values, you are comparing pointer values. If you want to compare string values you need to use a string comparison function such as strcmp.
if ( 0 == strcmp(var1, "dev")) {
..
}
"dev" is not a string it is a const char * like var1. Thus you are indeed comparing the memory adresses. Being that var1 is a char pointer, *var1 is a single char (the first character of the pointed to character sequence to be precise). You can't compare a char against a char pointer, which is why that did not work.
Being that this is tagged as c++, it would be sensible to use std::string instead of char pointers, which would make == work as expected. (You would just need to do const std::string var1 instead of const char *var1.
There is more stable function, also gets rid of string folding.
// Add to C++ source
bool string_equal (const char* arg0, const char* arg1)
{
/*
* This function wraps string comparison with string pointers
* (and also works around 'string folding', as I said).
* Converts pointers to std::string
* for make use of string equality operator (==).
* Parameters use 'const' for prevent possible object corruption.
*/
std::string var0 = (std::string) arg0;
std::string var1 = (std::string) arg1;
if (var0 == var1)
{
return true;
}
else
{
return false;
}
}
And add declaration to header
// Parameters use 'const' for prevent possible object corruption.
bool string_equal (const char* arg0, const char* arg1);
For usage, just place an 'string_equal' call as condition of if (or ternary) statement/block.
if (string_equal (var1, "dev"))
{
// It is equal, do what needed here.
}
else
{
// It is not equal, do what needed here (optional).
}
Source: sinatramultimedia/fl32 codec (it's written by myself)
your thinking about this program below
#include <stdio.h>
#include <string.h>
int main ()
{
char str[][5] = { "R2D2" , "C3PO" , "R2A6" };
int n;
puts ("Looking for R2 astromech droids...");
for (n=0 ; n<3 ; n++)
if (strncmp (str[n],"R2xx",2) == 0)
{
printf ("found %s\n",str[n]);
}
return 0;
}
//outputs:
//
//Looking for R2 astromech droids...
//found R2D2
//found R2A6
when you should be thinking about inputting something into an array & then use strcmp functions like the program above ... check out a modified program below
#include <iostream>
#include<cctype>
#include <string.h>
#include <string>
using namespace std;
int main()
{
int Students=2;
int Projects=3, Avg2=0, Sum2=0, SumT2=0, AvgT2=0, i=0, j=0;
int Grades[Students][Projects];
for(int j=0; j<=Projects-1; j++){
for(int i=0; i<=Students; i++) {
cout <<"Please give grade of student "<< j <<"in project "<< i << ":";
cin >> Grades[j][i];
}
Sum2 = Sum2 + Grades[i][j];
Avg2 = Sum2/Students;
}
SumT2 = SumT2 + Avg2;
AvgT2 = SumT2/Projects;
cout << "avg is : " << AvgT2 << " and sum : " << SumT2 << ":";
return 0;
}
change to string except it only reads 1 input and throws the rest out
maybe need two for loops and two pointers
#include <cstring>
#include <iostream>
#include <string>
#include <stdio.h>
using namespace std;
int main()
{
char name[100];
//string userInput[26];
int i=0, n=0, m=0;
cout<<"your name? ";
cin>>name;
cout<<"Hello "<<name<< endl;
char *ptr=name;
for (i = 0; i < 20; i++)
{
cout<<i<<" "<<ptr[i]<<" "<<(int)ptr[i]<<endl;
}
int length = 0;
while(name[length] != '\0')
{
length++;
}
for(n=0; n<4; n++)
{
if (strncmp(ptr, "snit", 4) == 0)
{
cout << "you found the snitch " << ptr[i];
}
}
cout<<name <<"is"<<length<<"chars long";
}
Is there anyway , if I enter any string , then I want to scan ASCII value of each character inside that string , if I enter "john" then I should get 4 variables getting ASCII value of each character, in C or C++
Given a string in C:
char s[] = "john";
or in C++:
std::string s = "john";
s[0] gives the numeric value of the first character, s[1] the second an so on.
If your computer uses an ASCII representation of characters (which it does, unless it's something very unusual), then these values are the ASCII codes. You can display these values numerically:
printf("%d", s[0]); // in C
std::cout << static_cast<int>(s[0]); // in C++
Being an integer type (char), you can also assign these values to variables and perform arithmetic on them, if that's what you want.
I'm not quite sure what you mean by "scan". If you're asking how to iterate over the string to process each character in turn, then in C it's:
for (char const * p = s; *p; ++p) {
// Do something with the character value *p
}
and in (modern) C++:
for (char c : s) {
// Do something with the character value c
}
If you're asking how to read the string as a line of input from the terminal, then in C it's
char s[SOME_SIZE_YOU_HOPE_IS_LARGE_ENOUGH];
fgets(s, sizeof s, stdin);
and in C++ it's
std::string s;
std::cin >> s; // if you want a single word
std::getline(std::cin, s); // if you want a whole line
If you mean something else by "scan", then please clarify.
You can simply get the ascii value of a char by casting it to type int:
char c = 'b';
int i = c; //i contains ascii value of char 'b'
Thus, in your example the code to get the ascii values of a string would look something like this:
#include <iostream>
#include <string>
using std::string;
using std::cout;
using std::endl;
int main()
{
string text = "John";
for (int i = 0; i < text.size(); i++)
{
cout << (int)text[i] << endl; //prints corresponding ascii values (one per line)
}
}
To get the corresponding char from an integer representing an entry in the ascii table, you just have to cast the int back to char again:
char c = (char)74 // c contains 'J'
The code given above was written in C++ but it basically works the same way in C (and many other languages as well I guess)
There is no way to turn a string of length 'x' into x variables. In C or C++ you can only declare a fixed number of variables. But probably you don't need to do what you are saying. Perhaps you just need an array, or most likely you just need a better way to solve whatever problem you are trying to solve. If you explain what the problem is in the first place, then I'm sure a better way can be explained.
Ya,I think there are some more better solutions are also available but this one also be helpful.
In C
#include <stdio.h>
#include <string.h>
#include <malloc.h>
int main(){
char s[]="abc";
int cnt=0;
while(1){
if(s[cnt++]==NULL)break;
}
int *a=(int *)malloc(sizeof(int)*cnt);
for(int i=0;i<cnt;i++)a[i]=s[i];
for(int i=0;i<cnt-1;i++)printf("%d\n",a[i]);
return 0;
}
In C++
#include <iostream>
#include <string>
using namespace std;
int main(){
string s="abc";
//int *a=new int[s.length()];
//for(int i=0;i<s.length();i++)a[i]=s[i];
for(int i=0;i<s.length();i++)
cout<<(int)s[i]<<endl;
return 0;
}
I hope this one will be helpful..
yeah it's very easy ..just a demo
int main()
{
char *s="hello";
while(*s!='\0')
{
printf("%c --> %d\n",*s,*s);
s++;
}
return 0;
}
But make sure your machine is supporting the ASCII value format.
In C every char has one integral value associted with it called ASCII.
Using %d format specifier you can directly print the ASCII of any char as above.
NOTE: It's better to get good book and practice this kind of program yourself.
I want to build a function to easily convert a string containing hex code (eg. "0ae34e") into a string containing the equivalent ascii values and vice versa.
Do I have to cut the Hex string in pairs of 2 values and gue them together again or is there a convenient way to do that?
thanks
Based on binascii_unhexlify() function from Python:
#include <cctype> // is*
int to_int(int c) {
if (not isxdigit(c)) return -1; // error: non-hexadecimal digit found
if (isdigit(c)) return c - '0';
if (isupper(c)) c = tolower(c);
return c - 'a' + 10;
}
template<class InputIterator, class OutputIterator> int
unhexlify(InputIterator first, InputIterator last, OutputIterator ascii) {
while (first != last) {
int top = to_int(*first++);
int bot = to_int(*first++);
if (top == -1 or bot == -1)
return -1; // error
*ascii++ = (top << 4) + bot;
}
return 0;
}
Example
#include <iostream>
int main() {
char hex[] = "7B5a7D";
size_t len = sizeof(hex) - 1; // strlen
char ascii[len/2+1];
ascii[len/2] = '\0';
if (unhexlify(hex, hex+len, ascii) < 0) return 1; // error
std::cout << hex << " -> " << ascii << std::endl;
}
Output
7B5a7D -> {Z}
An interesting quote from the comments in the source code:
While I was reading dozens of programs that encode or decode the
formats here (documentation? hihi:-) I have formulated Jansen's
Observation:
Programs that encode binary data in ASCII are written in such a style
that they are as unreadable as possible. Devices used include
unnecessary global variables, burying important tables in unrelated
sourcefiles, putting functions in include files, using
seemingly-descriptive variable names for different purposes, calls to
empty subroutines and a host of others.
I have attempted to break with this tradition, but I guess that that
does make the performance sub-optimal. Oh well, too bad...
Jack Jansen, CWI, July 1995.
If you want to use a more c++ native way, you can say
std::string str = "0x00f34" // for example
stringstream ss(str);
ss << hex;
int n;
ss >> n;
The sprintf and sscanf functions can already do that for you. This code is an example that should give you an idea. Please go through the function references and the safe alternatives before you use them
#include <stdio.h>
int main()
{
int i;
char str[80]={0};
char input[80]="0x01F1";
int output;
/* convert a hex input to integer in string */
printf ("Hex number: ");
scanf ("%x",&i);
sprintf (str,"%d",i,i);
printf("%s\n",str);
/* convert input in hex to integer in string */
sscanf(input,"%x",&output);
printf("%d\n",output);
}
When I use getline, I would input a bunch of strings or numbers, but I only want the while loop to output the "word" if it is not a number.
So is there any way to check if "word" is a number or not? I know I could use atoi() for
C-strings but how about for strings of the string class?
int main () {
stringstream ss (stringstream::in | stringstream::out);
string word;
string str;
getline(cin,str);
ss<<str;
while(ss>>word)
{
//if( )
cout<<word<<endl;
}
}
Another version...
Use strtol, wrapping it inside a simple function to hide its complexity :
inline bool isInteger(const std::string & s)
{
if(s.empty() || ((!isdigit(s[0])) && (s[0] != '-') && (s[0] != '+'))) return false;
char * p;
strtol(s.c_str(), &p, 10);
return (*p == 0);
}
Why strtol ?
As far as I love C++, sometimes the C API is the best answer as far as I am concerned:
using exceptions is overkill for a test that is authorized to fail
the temporary stream object creation by the lexical cast is overkill and over-inefficient when the C standard library has a little known dedicated function that does the job.
How does it work ?
strtol seems quite raw at first glance, so an explanation will make the code simpler to read :
strtol will parse the string, stopping at the first character that cannot be considered part of an integer. If you provide p (as I did above), it sets p right at this first non-integer character.
My reasoning is that if p is not set to the end of the string (the 0 character), then there is a non-integer character in the string s, meaning s is not a correct integer.
The first tests are there to eliminate corner cases (leading spaces, empty string, etc.).
This function should be, of course, customized to your needs (are leading spaces an error? etc.).
Sources :
See the description of strtol at: http://en.cppreference.com/w/cpp/string/byte/strtol.
See, too, the description of strtol's sister functions (strtod, strtoul, etc.).
The accepted answer will give a false positive if the input is a number plus text, because "stol" will convert the firsts digits and ignore the rest.
I like the following version the most, since it's a nice one-liner that doesn't need to define a function and you can just copy and paste wherever you need it.
#include <string>
...
std::string s;
bool has_only_digits = (s.find_first_not_of( "0123456789" ) == std::string::npos);
EDIT: if you like this implementation but you do want to use it as a function, then this should do:
bool has_only_digits(const string s){
return s.find_first_not_of( "0123456789" ) == string::npos;
}
You might try boost::lexical_cast. It throws an bad_lexical_cast exception if it fails.
In your case:
int number;
try
{
number = boost::lexical_cast<int>(word);
}
catch(boost::bad_lexical_cast& e)
{
std::cout << word << "isn't a number" << std::endl;
}
If you're just checking if word is a number, that's not too hard:
#include <ctype.h>
...
string word;
bool isNumber = true;
for(string::const_iterator k = word.begin(); k != word.end(); ++k)
isNumber &&= isdigit(*k);
Optimize as desired.
Use the all-powerful C stdio/string functions:
int dummy_int;
int scan_value = std::sscanf( some_string.c_str(), "%d", &dummy_int);
if (scan_value == 0)
// does not start with integer
else
// starts with integer
You can use boost::lexical_cast, as suggested, but if you have any prior knowledge about the strings (i.e. that if a string contains an integer literal it won't have any leading space, or that integers are never written with exponents), then rolling your own function should be both more efficient, and not particularly difficult.
Ok, the way I see it you have 3 options.
1: If you simply wish to check whether the number is an integer, and don't care about converting it, but simply wish to keep it as a string and don't care about potential overflows, checking whether it matches a regex for an integer would be ideal here.
2: You can use boost::lexical_cast and then catch a potential boost::bad_lexical_cast exception to see if the conversion failed. This would work well if you can use boost and if failing the conversion is an exceptional condition.
3: Roll your own function similar to lexical_cast that checks the conversion and returns true/false depending on whether it's successful or not. This would work in case 1 & 2 doesn't fit your requirements.
Here is another solution.
try
{
(void) std::stoi(myString); //cast to void to ignore the return value
//Success! myString contained an integer
}
catch (const std::logic_error &e)
{
//Failure! myString did not contain an integer
}
Since C++11 you can make use of std::all_of and ::isdigit:
#include <algorithm>
#include <cctype>
#include <iostream>
#include <string_view>
int main([[maybe_unused]] int argc, [[maybe_unused]] char *argv[])
{
auto isInt = [](std::string_view str) -> bool {
return std::all_of(str.cbegin(), str.cend(), ::isdigit);
};
for(auto &test : {"abc", "123abc", "123.0", "+123", "-123", "123"}) {
std::cout << "Is '" << test << "' numeric? "
<< (isInt(test) ? "true" : "false") << std::endl;
}
return 0;
}
Check out the result with Godbolt.
template <typename T>
const T to(const string& sval)
{
T val;
stringstream ss;
ss << sval;
ss >> val;
if(ss.fail())
throw runtime_error((string)typeid(T).name() + " type wanted: " + sval);
return val;
}
And then you can use it like that:
double d = to<double>("4.3");
or
int i = to<int>("4123");
I have modified paercebal's method to meet my needs:
typedef std::string String;
bool isInt(const String& s, int base){
if(s.empty() || std::isspace(s[0])) return false ;
char * p ;
strtol(s.c_str(), &p, base) ;
return (*p == 0) ;
}
bool isPositiveInt(const String& s, int base){
if(s.empty() || std::isspace(s[0]) || s[0]=='-') return false ;
char * p ;
strtol(s.c_str(), &p, base) ;
return (*p == 0) ;
}
bool isNegativeInt(const String& s, int base){
if(s.empty() || std::isspace(s[0]) || s[0]!='-') return false ;
char * p ;
strtol(s.c_str(), &p, base) ;
return (*p == 0) ;
}
Note:
You can check for various bases (binary, oct, hex and others)
Make sure you don't pass 1, negative value or value >36 as base.
If you pass 0 as the base, it will auto detect the base i.e for a string starting with 0x will be treated as hex and string starting with 0 will be treated as oct. The characters are case-insensitive.
Any white space in string will make it return false.