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I've been trying to learn C programming by reading a textbook, but am confused about how strings and substrings work.
I have an idea of what strings and substrings are from java, but can't figure out the syntax in C.
Here's a question from the book that I thought might be easy, but I can't get it.
Write and test a function hydroxide that returns a 1 for true if its string argument ends in the substring OH.
It recommends testing the function with KOH and NaCl.
Also, how would I remove and add letters at the end of the string?
Like, if for some reason I wanted to change NaCl to NaOH?
Any help and explanations would be really appreciated.
ETA:
I guess what I'm most confused on is how to make the program look at the last two letters in the string and compared them to OH.
I'm also not sure how to pass strings to functions.
String is a sequence of characters that ends with special null-terminated character '\0'. If there is no \0, functions that work with string won't stop until the \0 symbol is found. This character may happen in any place after the end of pseudo string (I mean a string without \0) and only then stop.
The following example shows the necessity of this null-terminated character:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char string[] = "Hello!";
printf("original string:\n%s\n\n", string);
memset(string, '-', 5);
printf("memset doesn't affect the last two symbols: '!' and '\\0':\n%s", string);
memset(string, '-', 6);
printf("\n\nmemset doesn't affect the last symbol: '\\0':\n%s\n\n", string);
memset(string, '-', 7);
printf("memset affects all symbols including null-terminated one:\n%s", string);
return 0;
}
/* OUTPUT:
original string:
Hello!
memset doesn't affect the last two characters: '!' and '\0':
-----!
memset doesn't affect the last character: '\0':
------
memset affects all characters including null-terminated one:
-------#↓#
*/
Substring is a char sequence that is in a string. It may be less or equal to the string.
Suppose, "NaOH" is a string. Then substring may be: "N", "a", "O", "H", "Na", "aO", "OH", "NaO", "aOH", "NaOH". To find whether substring is in the string or not you can use strstr function. It's prototype is char * strstr ( char * str1, const char * str2 );.
This code shows this function's results:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *ptrCh = NULL;
ptrCh = strstr("hello", "h");
printf("ptrCh: %p\n", ptrCh);
printf("%s\n\n", ptrCh);
ptrCh = strstr("hello", "z");
printf("ptrCh: %p\n", ptrCh);
printf("%s\n\n", ptrCh);
return 0;
}
/* OUTPUT:
ptrCh: 00403024
hello
ptrCh: 00000000
(null)
*/
As for the first printf, it prints characters beginning from the position of 'h' and when it reaches null-terminated character, which is next after 'o', it stops, exactly as in previous program.
To make your program more interactive, you can declare array and then a pointer to it. Array size must be enough to store the longest formula. Suppose, 100 will be enough:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char buf[100] = {0};
char *ptr = &buf[0];
scanf("%s", ptr);
// printf() gets a pointer as argument
printf("%s\n", ptr);
// printf() gets also a pointer as argument.
// When you pass arrays name without index to a function,
// you pass a pointer to array's first element.
printf("%s", buf);
return 0;
}
And as for rewriting letters in the end of the string. Here is a small program that does it. Pay attention at comments:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char buf[100] = {0};
char formula[100] = {0};
char compound[100] = {0};
char *ptr = &buf[0];
char *pFormula = &formula[0];
char *pCompound = &compound[0];
printf("Enter formula: ");
scanf("%s", pFormula);
printf("Enter chemical compound: ");
scanf("%s", pCompound);
// Copying the first chemical elements without the last
// several that will be replaced by another elements.
strncpy(ptr, pFormula, strlen(pFormula) - strlen(pCompound));
// Adding new compound to the first elements.
// Function also adds a null-terminated character to the end.
strncat(ptr, pCompound, strlen(pCompound));
printf("The new chemical compound is: ");
printf("%s", ptr);
return 0;
}
/* OUTPUT:
Enter formula: NaOH
Enter chemical compound: Cl
The new chemical compound is: NaCl
*/
In C, we use null-terminated strings. That is the "invisible", 0 value. Not ASCII "0", but the zero value, like 8-bit 0x00. You can represent it in literal text with '\0' or "\0" or unquoted 0, however, in a literal string it is redundant because most functions like strcmp() or strstr() or strcat() all expect and work with null terminated strings. Null char is the stops sign for the C standard string functions.
One easy way to implement this with C library calls is to test for existence of the substring and then test that substring's length, which verify it is at end of string.
Assume buf is some big string buffer, char buf[1024] and char *temp is a temporary variable.
temp = strstr(buf, "OH") returns the pointer to "OH" if exists in buf at any offset.
strlen(temp) Get length of temp, if at end of string, it will be 2 (doesn't include null terminator), so if the original string is "OHIO" or "SOHO" it wont match because it'll be 4 and 3 respectively.
The above is the core of the code, not the full robust implementation. You need to check for valid return values, etc.
char buf[1024];
char *temp;
strcpy(buf, "NaOH");
if((temp = strstr(buf, "OH")) != 0)
{
// At this point we know temp points to something that starts with "OH"
// Now see if it is at the end of the string
if(strlen(temp) == 2)
return true; // For C99 include stdbool.h
return false;
}
You could get obscure, and check for the null terminator directly, will be a smidge quicker. This code is safe as long as it is inside the if() for strstr(), otherwise never do this if you don't know a string is a least N characters long.
if(temp[2] == '\0')
return true; // For C99 include stdbool.h
As far as appending to a string, read the docs on strcat. Keep in mind with strcat, you must have enough space already in the buffer you are appending into. It isn't like C++ std::string or Java/C# string where those will dynamically resize as needed. In C, you get to do all of that yourself.
Related
I am trying to reverse a string in C with pointers but the output is really weird. The logic seems good but I am not sure why it gets outputted like this. Here is the code:
#include <stdio.h>
#include <string.h>
int main()
{
char str[20], reverse_str[20], *pointer;
int i = 0;
printf("%s", "Enter any string: ");
scanf("%s", str);
pointer = str;
int string_length = strlen(pointer);
//int count = 0;
for (int i = string_length; i > 0; i--){
reverse_str[i -1] = *pointer;
pointer++;
}
printf("%d\n", string_length);
printf("Original string = %s\n", str);
printf("Reversed string = %s\n", reverse_str);
}
The output looks like this:
Enter any string: Hello
Original string = Hello
Reversed string = olleH╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠Hello
This is when the role of null terminator comes in which is often ignored as a silly bit of information by beginners.
In C, every string is terminated by a null character because there needs to be some way to know when a particular string ends starting from its initial location in memory. If you properly initialize the string or put in \0 appropriately then the string can be displayed as is. Otherwise every character in memory starting from the zeroth index of the string will be displayed until it encounters a null character. This is how a printf("%s", str) works, in simple words.
You get that weird output because of this reason. This explains it much better.
Solution:
Add reverse_str[string_length] = '\0'; after you reverse your string using that for loop, so that your resultant string is properly null terminated.
Bonus:
And the reason why you got a considerably sane output is that you were lucky since the compiler allocated str and reverse_str close to each other in a direction such that even if you miss the null terminator on reverse_str you hit the null terminator of str.
To print a string, the string needs a NUL-terminator \0 but reverse_str doesn´t have one.
You need to set a \0 at the end of revers_str, like:
reverse_str[string_length] = '\0';
to make it print the right output.
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When I try to convert a char[] to char* there are unneeded characters added to the char* variable
int keySize = getKeySize(key2);
char* key = (char*)malloc(sizeof(keySize));
int i;
char s[keySize-1];
int i2;
for(i2=0; i2<keySize; i2++)
{
s[i2] = getCharacter(key2, i2);
}
strncpy(key, s, keySize);
There is no string data type in C programming language. Strings in C are represented as array of characters.
Note: C-Strings are actually character array terminated by '\0' character. That means, last character in any C-String in C will be used to store a '\0' character which marks the end of the string. To store n characters in C-String in C, you should define a character array of size (n+1).
Why should we terminated it by '\0'?
The '\0' termination is what differentiates a char array from a c-string in C programming language. Most string-manipulating functions (like strcpy) relies on '\0' character to know when the string is finished (and its job is done!), and won't work with simple char-array (eg. they'll keep on working past the boundaries of the array, and continue until it finds a '\0' character somewhere in memory - often corrupting memory as it goes).
Therefore, storing a '\0' character (at the end) is necessary if you want to use functions of #include <string.h> like strcpy() as they rely on '\0' character to mark the end of the character array.
'\0' is defined to be a null character - that is a character with all bits set to zero (and thus has a value 0). This has nothing to do with pointers. Read more about it here.
In your program, you want two character arrays key (dynamically allocated) and s to hold a copy of another character array key2 of size keysize. Then, both character arrays should be of atleast keysize + 1 (+1 to hold a '\0' character) size.
Change:
char* key = (char*)malloc(sizeof(keySize));
To:
char* key = malloc(keySize+1); // Don't Type-Cast malloc
And
Change:
char s[keySize-1];
To
char s[keySize+1];
While allocating, you should allocate one more than the size. Currently you are allocating 4 bytes only.
char* key = (char*)malloc(keySize+1);
//instead of
char* key = (char*)malloc(sizeof(keySize));
s should have a size of keySize+1
char s[keySize+1];
// instead of
char s[keySize-1];
What about this?
There are some errors about the dimension of s, I suggest you tu use strncpy
#include <string.h>
int main(){
//bla bla ...
int keySize = getKeySize(key2);
char* key = malloc(keySize+1);;
int i2;
for(i2=0; i2<keySize; i2++){
s[i2] = getCharacter(key2, i2);
}
char s[keySize+1];
strncpy(s, key, sizeof s - 1);
s[keySize] = '\0';
r
return 0;
}
Anyway more information about it please,I supposed you wanted this
Anyone know how to copy to strings? Cause I used the function strcpy but when I print the result it show strange characters. I want to concatenate 'name' + '#' + 'e-mail'. With scanf I have to put the character null '\0'?
#include <stdio.h>
#include <string.h>
int main (){
char message[150];
char name[150];
char mail[150];
char result[150];
printf("Introduce name: \n");
scanf("%s",message);
printf("Introduce email \n");
scanf("%s",server);
strcpy(result,message);
result[strlen(result)]='#';
strcpy(&result[strlen(result)],server);
printf("RESULT: %s\n",result);
return 0;
}
result[strlen(result)]='#'; overwrites the NUL terminator introduced into result by strcpy(result,message);. So the result of a subsequent strlen is undefined.
A better solution is to use strncat, or you could get away with writing
char result[150] = {'\0'};
which will initialise the entire array.
But you still run the risk of overflowing your result array. You could use the safer strncpy to obviate that. Better still, use snprintf and have the C standard library perform the concatenation for you.
reverser() reverses a cstring (not in place). 99% of the time it works but some input corrupts it for example it appears if aStr2[] is assigned a string made up of the same character it will have an error.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* reverser(const char *str);
int main()
{
char aStr[] = "aaa";
char aStr2[] = "cccccc";
printf("%s %s", aStr, aStr2);
char* tmp = reverser(aStr2);//tmp now has garbage
printf("\n%s", tmp);
printf(" %s", aStr2);
return 0;
}
char* reverser(const char *str)
{
char* revStr = (char*)malloc(strlen(str));
int i;
for(i = strlen(str)-1; i >= 0; i--)
{
revStr[strlen(str)-1-i] = str[i];
}
return revStr;
}
Gives
aaa cccccc
cccccc9 cccccc
Process returned 0 (0x0) execution time : 0.068 s
Press any key to continue
Notice the 9 that shouldn't be there.
Change this malloc to strlen(str) + 1 , plus 1 for '\0'
char* revStr = (char*)malloc(strlen(str) + 1);
and after the for loop
revStr[strlen(str)+1] = '\0';
Your problem is that you don't put the string terminator in your reversed string. All strings in C are actually one extra character that isn't reported by strlen, and that is the character '\0' (or plain and simple, a zero). This tells all C functions when the string ends.
Therefore you need to allocate space for this extra terminator character in your malloc call, and add it after the last character in the string.
There are also a couple of other problems with your code, the first is that you should not cast the return of malloc (or any other function returning void *). Another that you have a memory leak in that you do not free the memory you allocate. This last point doesn't matter in a small program like the one you have here, but will be an issue in larger and longer running programs.
You haven't null-terminated your reversed string. You need to set the final index of revStr[] to 0.
I have this example code for the strchr function in C.
/* strchr example */
#include <stdio.h>
#include <string.h>
int main ()
{
char str[] = "This is a sample string";
char * pch;
printf ("Looking for the 's' character in \"%s\"...\n",str);
pch=strchr(str,'s');
while (pch!=NULL)
{
printf ("found at %d\n",pch-str+1);
pch=strchr(pch+1,'s');
}
return 0;
}
The problem is, I don't understand, how this program calculates the position of the looking character.
I think it has something to do with the pointers of "pch" and "str", but how does this work?
Would be great, if there is somebody who could explain this in little more detail.
thanks,
eljobso
It simply subtracts str, which is a pointer to the first character of the string, from the pointer to the found result.
This then becomes the position of the character, indexed from 0. This is easy to understand, if the character is found in the first position of the string, the returned pointer will be equal to str, and thus (pstr - str) == 0 is true. Adding one makes it 1-based, which is sometimes useful for presentational purposes.