*char conversion to/from bool in call to strstr - c++

I have code that is supposed to remove all characters in one C-string from another.
The problem arises when I try to use the function strstr: both an array and a char* get converted to bool. Obviously it doesn't work because strstr needs to receive 2 char* arguments.
#include <iostream>
#include <ctype.h>
#include <string.h>
#include <malloc.h>
using namespace std;
char* temp_char(char* orig,char* del)
{ char *str_temp, *temp, *p, *c;
char symbol[2];
int i=0;
int len = strlen(orig);
temp=(char *)calloc(len,1);
str_temp = (char *)calloc(len,1);
strcpy(str_temp,orig);
for(i=0;i<strlen(del);i++){
symbol[0]=del[i];
symbol[1]=0;
temp[0]=0;
while( p=strstr(str_temp,del)!=NULL){
strncat(temp,str_temp,p-str_temp);
p++;
str_temp=p;
}
strcat(temp,str_temp);
strcpy(str_temp,temp);
}
cout<<temp;
return temp;
}
int main () {
char a[]="stuccsess",b[]="test";
temp_char(a,b);}
Any help would be appreciated.

You have two errors in one line of your code. The first is not addressing the issue that the != operator has higher priority than =. Thus, in the following line:
while( p=strstr(str_temp,del)!=NULL) {
the comparison is actually:
while( p = ( strstr(str_temp,del)!=NULL ) ){
So, you are attempting to assign the result of the != test, which will be a bool value, to a char* variable (p).
You can fix this, easily, by putting the assignment expression in parentheses.
However, although that will fix the compiler error, it will not address the second error. Just a few lines above that while statement, you assign one of the characters of the del string to the first of the symbol string (and, properly, add a nul terminator to that) … but you never then actually use that symbol string. Instead, you pass the entire del string as the second argument in the call to strstr. You should be passing your created symbol as that second argument.
So, changing that while line to the following will make the code work:
while ((p = strstr(str_temp, symbol)) != nullptr) { // Use nullptr in C++
But that leaves other errors in your function. How will you ever be able to free the memory allocated in the str_temp = (char*)calloc(len, 1); call? Once you have subsequently modified that str_temp pointer (in the str_temp = p; line), you no longer have the original pointer value, which must be used when calling free(). So, you need to save that value, just after you have made the allocation. (The temp memory pointer is returned, so that can be freed in the main function.)
There are other issues in your code that could be improved, like using new[] and delete[] in C++, rather than the old, C-language calloc, and that your index and length variables should really be of size_t type, rather than int. Here's a version with the corrections and suggestions discussed applied:
#include <iostream>
#include <cstring>
using std::cout;
char* temp_char(char* orig, char* del)
{
size_t len = strlen(orig);
char* temp = new char[len + 1];
char* str_temp = new char[len + 1];
char* save_temp = temp; // Save it so we can call delete...
strcpy(str_temp, orig);
for (size_t i = 0; i < strlen(del); i++) {
char symbol[2];
symbol[0] = del[i];
symbol[1] = 0;
temp[0] = 0;
char* p;
while ((p = strstr(str_temp, symbol)) != nullptr) {
strncat(temp, str_temp, static_cast<size_t>(p - str_temp));
p++;
str_temp = p;
}
strcat(temp, str_temp);
strcpy(str_temp, temp);
}
cout << temp;
delete[] save_temp; // ... don't forget to free this.
return temp;
}
int main()
{
char a[] = "stuccsess", b[] = "test";
char* answer = temp_char(a, b);
delete[] answer; // We should free the memory allocated!
return 0;
}
But your approach is far more complicated than it need be. You can simply loop through the original string and check each character to see if it is in the 'delete' string (using the strchr function, for example); if it is not (i.e. that strchr returns nullptr), then append the character to the accumulated temp and increase the running length:
char* temp_char(char* orig, char* del)
{
size_t len = strlen(orig);
char* temp = new char[len + 1];
int newlen = 0;
for (char* cp = orig; *cp != '\0'; ++cp) {
if (strchr(del, *cp) == nullptr) temp[newlen++] = *cp;
}
temp[newlen] = '\0'; // Add null terminator
std::cout << temp;
return temp;
}

Related

Copying char* with char symbol

I am trying to copy a char* and char symbol into a new char* , however the desired result after copying is wrong.
char* name = "someData";
char symbol ='!';
int size1 = strlen(name);
int size2 = 1;
int newSize = size1 + size2 + 1;
char* res = new char[newSize];
strcpy(res,name);
const char* symbolPointer = &symbol;
strcat(res, symbolPointer);
cout<<*res;
I expect the result to be "someData!" , however it is only "s" , where is my mistake?
char* name = "someData";
This is an ill-formed conversion in C++ (since C++11). I recommend to not point to string literals with pointer to non-const.
const char* symbolPointer = &symbol;
strcat(res, symbolPointer);
Both arguments of std::strcat must be null terminated. symbolPointer is not a pointer to a null terminated string. Because the pre-condition of std::strcat is violated, the behaviour of the program is undefined.
cout<<*res;
res is a pointer to the first character of the string. By indirecting through the pointer to first character, you get the first character. That is why you see the first character (in case the undefined behaviour hasn't caused the program to do something completely different).
Lastly, the program leaks the allocated res.
Here is a fixed example:
std::string name = "someData";
name += '!';
std::cout << name;
The problem is the following:
cout<<*res;
This is equivalent to:
cout << res[0];
It prints just the first character of the output. Use
cout<<res;
Try it this way:
const string name = "someData";
const char symbol ='!';
string res = name + symbol;
cout << res;
You should avoid the legacy C nul-terminated string handling functions. You should avoid using new directly in your code.
The observed result you are asking about is due to you writing *res (a single character) instead of res (a pointer to the first character) in the output statement. But the code was buggy besides that, as strcat will copy until it finds the terminator, so it will overwrite some unknown amount of memory beyond what you allocated.
symbolPointer should end with null character, because strcat requires 0-terminated string.
So if you want to continue on your way,
(not a good idea but)
You can add this before strcat.
*(symbolPointer+1) = 0;
#include <iostream>
#include <cstring>
int main() {
char name[] = "someData";
char symbol ='!';
int size1 = strlen(name);
int size2 = 1;
int newSize = size1 + size2 + 1;
char* res = new char[newSize];
strcpy(res,name);
char* symbolPointer = &symbol;
*(symbolPointer + 1) = 0;
strcat(res, symbolPointer);
cout<<res;
return 0;
}

C++ Junior interview question: function to compress a character sequence with only char pointers

I was at a job interview the other day and I had the following function to implement:
char* Compress (char * text);
Rules were also that you are not allowed to use standard libary functions like strlen, strcpy, string etc... So the function has to compress a given character sequence.
For example if the input text is"11112222333344411" after passing it to the Compress function the returned value is: "12341", or if the text input is:"aaAbbBBcCCa" ---> return: aAbBcCa
I am not sure I did everything properly (with memory handling) here so any suggestions would be great. Am I doing it right that I delete the value of temp every time? Also if there is a simpler way to implement this function (without using the standard library functions of course) I would be really pleased to see it.
#include <iostream>
char* Compress(char* text) {
char* temp;
char* _compText;
int size = 1;
_compText = nullptr;
for (size_t i = 0; text[i] != '\0'; ++i)
{
if (text[i] != text[i + 1]) {
++size;
temp = _compText;
_compText = new char[size];
for (size_t j = 0; j < size-2; ++j)
{
_compText[j] = temp[j];
}
_compText[size-2] = text[i];
_compText[size-1] = '\0';
delete[] temp;
}
}
return _compText;
}
int main()
{
char t[] = "111122222233333444444555555111";
char* compedT;
std::cout << "Before:\n";
std::cout << t;
compedT = Compress(t);
std::cout << "\nAfter: \n";
std::cout << compedT;
delete[] compedT;
return 0;
}
The function initially is implemented incorrectly.
The type of the function is
char* Compress (char * text);
^^^^^^^
that is its parameter is not const char *, This means that the function should update the source string in place and return pointer to its first character. There is no need to allocate dynamically memory to perform the task.
The function can be defined as it is shown in the demonstrative program.
#include <iostream>
char * Compress( char *s )
{
for ( char *p = s, *q = s; *q; )
{
if ( *++q != *p ) *++p = *q;
}
return s;
}
int main()
{
char s[] = "11112222333344411";
std::cout << Compress( s ) << '\n';
}
Its output is
12341
Or the function can look also the following way
char * Compress( char *s )
{
for ( char *p = s, *q = s; *q; )
{
if ( ( *++q != *p ) and ( ++p != q ) ) *p = *q;
}
return s;
}
As for your function implementation then you should read warnings as for example
warning: comparison of integer expressions of different signedness: 'size_t' {aka 'long unsigned int'} and 'int' [-Wsign-compare]
34 | for (size_t j = 0; j < size-2; ++j)
| ~~^~~~~~~~
And your function returns nullptr for an empty string. This looks logically inconsistent. And the function is very inefficient.:)
And do not use names that start with underscore.
Does my code have memory leak?
As far as I can see, no; there is no memory leak.
That said, the use of bare owning pointers makes it difficult to spot memory leaks. They are a bad design choice, especially when transferring ownership to outside of the function. At the very least, there should be a comment near the function declaration that should document how the caller must clean up the allocation. If you need a dynamic array, a better solution is to use a container.
Am I doing it right that I delete the value of temp everytime?
As far memory leaks are concerned yes, you do need to delete every allocation. But reallocating memory on every iteration is unnecessary, and quite slow. In fact, there doesn't appear to be need for any dynamic allocation (See Vlad's answer).

Copying a C-style string into memory allocated on free store?

I'm doing an exercise in which I have to copy a c-style string into memory allocated on free store. I am required to do it without using subscripting and relying solely on pointer arithmetic. I wrote the following function-
char* str_dup(const char* s)
{
// count no. of elements
int i = 0;
const char* q = s;
while (*q) { ++i; ++q; }
//create an array +1 for terminating 0
char* scpy = new char[i + 1];
//copy elements to new array
while (*s)
{
*scpy = *s;
++s;
++scpy;
}
*scpy = 0;
return scpy;
}
The function is returning random characters. But if I change it into this-
char* str_dup(const char* s)
{
// count no. of elements
int i = 0;
const char* q = s;
while (*q) { ++i; ++q; }
//create an array +1 for terminating 0
char* scpyx = new char[i + 1];
char* scpy = scpyx;
//copy elements to new array
while (*s)
{
*scpy = *s;
++s;
++scpy;
}
*scpy = 0;
return scpyx;
}
it works. Can someone explain me why first code is not working and second is working?
The first code is not working since you return the final value of scpy, which at that point points at the terminating NUL character, and not the start of the string.
One solution is to do as you did, and save a copy of the original pointer to have something to return.
You should really use strlen() and memcpy(), they make this easier but perhaps they're off-limits to you.

C++ Combining two zero terminated strings?

So I am doing a question where I have to join two zero terminated strings, the first contains a word, and the second is empty and twice the size of the original array. I was able to get this working using the following code
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
char str1[] = "test";
char str2[(sizeof(str1)-1)*2];
char *p;
int count = 0;
for(p = str1; *p != 0; p++) {
str2[count] = *p;
count++;
}
cout << str2;
}
However I have to use a function with the following prototype
char *combine(char *a);
So I tried this
#include <stdio.h>
#include <iostream>
using namespace std;
char *copy_and_reverse(char *a) {
char str2[8];
int count = 0;
char* b = str2;
for(a; *a != 0; a++) {
str2[count] = *a;
count++;
}
return b;
}
int main()
{
char str1[] = "test";
char *a;
a = str1;
char* b = copy_and_reverse(a);
for(b; *b != 0; b++) {
cout << *b;
}
}
But it does not work (it is printing the string but it's printing a few random characters after it), I'm getting so confused with the pointers, can anyone help me out with this?
Edit: here is the question I am trying to answer
Write a function in C++ that takes as a char * style zero terminated string and returns a char* string twice the length of the input. The first half of the returned string should contain a copy of the contents of the original array. The second half of the string should contain the contents of the original string in reverse order.
The function should have the following prototype:
char *copy_and_reverse(char* a);
Note: you should not use any library functions (e.g from string.h).
There are two big problems in your copy_and_reverse code.
After copying the input string, you are not terminating the result. This means str2 is not a valid string. Fix:
str2[count] = '\0'; // after the loop
copy_and_reverse returns a pointer to a local variable (str2). After the function returns, all its local variables are gone, and main is dealing with an invalid pointer. To fix this, either use static memory (e.g. by declaring str2 as static or making it a global variable) or dynamic memory (allocate storage with new[] (or malloc())). Both approaches have their disadvantages.
Minor stuff:
variable; does nothing (see for (a; ...), for (b; ...)).
str2 isn't big enough for the final result. str1 is 5 bytes long ('t', 'e', 's', 't', '\0'), so char str2[8] is sufficient for now, but in the end you want to allocate length * 2 + 1 bytes for your result.
I believe that this will suit your needs:
#include <stdio.h>
#include <stdlib.h>
static char* copy_and_reverse(char* a);
static int strlen(char *c); // self-implemented
int main(void) {
char *a = "some string";
char *b = copy_and_reverse(a);
printf("%s", b);
free(b);
return 0;
}
static char* copy_and_reverse(char* a) {
int n = strlen(a);
char *b = new char[n * 2 + 1]; // get twice the length of a and one more for \0
for (int i = 0; i < n; ++i) { // does copying and reversing
b[i] = a[i];
b[i+n] = a[n-i-1];
}
b[2 * n] = '\0'; // null out last one
return b;
}
static int strlen(char *c) {
char *s = c;
while( *s++ );
return s-c-1;
}

Program crashes in the second run, after the return 0 from main C++

I have built my own functions of strlen and strdup.
When i use my strdup in the first time it's okay, i close the window, run it again, then in the end of the program after the return 0 from the main the program crashes. VS just says that it triggered a breakpoint.
#include "stdafx.h"
#include <iostream>
using namespace std;
int MyStrlen(const char* str);
char* MyStrdup(const char* str);
int main()
{
char *s1 = "Hello World!";
char *s2 = MyStrdup(s1);
cout << s1 << " , " << s2 << endl;
system("pause");
return 0;
}
int MyStrlen(const char* str)
{
register int iLength = 0;
while (str[iLength] != NULL)
{
iLength++;
}
return iLength;
}
char* MyStrdup(const char* str)
{
char* newStr;
int strLength = MyStrlen(str);
newStr = new char(strLength+1);
for (register int i = 0; i < strLength; i++)
{
newStr[i] = str[i];
}
newStr[strLength] = NULL;
return newStr;
}
Can someone note the place that makes it crash? I think it's a memory leak maybe.
Also, can you note things to improve in the code? For my learning purpose
EDIT: Thanks, I don't know why I used () instead of [] to define my new char[]. That was a memory leak or overwrite after all.
The "new" statement for an array should be with square brackets:
newStr = new char[strLength+1];
When you do
new char(c)
It allocates a single character and copies the character c into it.
When you do
new char[n]
it allocates memory for n characters
The expression new char(strLength+1) allocates a single character, and initializes it to strLength + 1. That of course means you will write out of bounds and have undefined behavior when you copy the string.
You should use new char[strLength + 1] instead, to allocate an "array" of characters.
On an unrelated note, while the terminating character in a string is commonly called the null character, it's not actually a null pointer (which is what NULL is for). Not that it really matters since in C++ NULL is a macro that expands to 0, but you should probably be explicit and use '\0' anyway (it gives more context for future readers).