Trying to do Exercise 1-19 of K&R 2nd ed., e.g. writing a function to reverse a string. I thought I managed, but the print output looks strange :-) If I use STRINGSIZE 5 the output is
Original String: hello
Reversed String: ollehhello. If I use STRINGSIZE 6 to keep in mind the '\0' string end character and modify the while-loop to while ((outputString[STRINGSIZE - (i + 2)] = inputString[i]) != '\0'), then I get Original String: hello
Reversed String: olleh?hello, I guess the ? is some random character coming from a '\0' added to the reversed string in the while-loop at position 5; but hello is again added. Could anyone explain how come hello gets added to the end of olleh and how can I get rid of it so that I only get the proper reversed string ?
Here is the code:
#include <stdio.h>
#define STRINGSIZE 5
void reverseString (char inputString[], char outputString[]);
int main(void) {
char stringToReverse[] = "hello";
char reversedString[STRINGSIZE];
reverseString(stringToReverse, reversedString);
printf("Original String: %s\nReversed String: %s\n", stringToReverse, reversedString);
}
void reverseString (char inputString[], char outputString[]) {
int i;
i = 0;
while ((outputString[STRINGSIZE - (i + 1)] = inputString[i]) != '\0')
++i;
}
First, the character array reversedString[] does not have enough space to store the null terminator of the string "hello". One option is to use a variable length array here:
char reversedString[strlen(stringToReverse) + 1];
VLAs were introduced in C99, and made optional in C11. As I remember it, K&R does not include coverage of variable length arrays, since even the 2nd edition was published before this.
Another option that would be compatible with C89 is to use the sizeof operator:
char stringToReverse[] = "hello";
char reversedString[sizeof stringToReverse];
Here, the result from the sizeof operator is known at compile time, and can be used in the declaration of a fixed size array. This size includes space for the null terminator, in contrast to the result from strlen("hello"). Note that this would not work with char *stringToReverse = "hello";, since then the sizeof operator would give the size of a pointer. This also would not work if stringToReverse had been passed into a function first, since then the array name would have decayed to a pointer to the first element of stringToReverse.
In the reverseString() function, the length of inputString needs to be determined (since STRINGSIZE is no longer being used); this can be done with strlen(), or in a loop. Then, critically, the function must be certain to add a null terminator (\0) to outputString[] before returning. Also note that a return statement has been added to the end of main() to make this truly C89 compatible:
#include <stdio.h>
void reverseString (char inputString[], char outputString[]);
int main(void) {
char stringToReverse[] = "hello";
char reversedString[sizeof stringToReverse];
reverseString(stringToReverse, reversedString);
printf("Original String: %s\nReversed String: %s\n",
stringToReverse, reversedString);
return 0;
}
void reverseString(char inputString[], char outputString[])
{
int length = 0;
int i = 0;
/* Get inputString length; or use strlen() */
while (inputString[length] != '\0') {
++length;
}
/* Copy to outputString[] in reverse */
while (i < length) {
outputString[i] = inputString[(length - i) - 1];
++i;
}
/* Add null terminator */
outputString[i] = '\0';
}
First I'll suggest that you change this line:
char reversedString[STRINGSIZE];
to
char reversedString[strlen(stringToReverse) + 1]; // + 1 to make room for the string termination
Then I would do something like:
void reverseString (char inputString[], char outputString[]) {
int i;
int len = strlen(inputString);
for(i=0; i<len; ++i)
{
outputString[len-i-1] = inputString[i];
}
outputString[len] = '\0'; // Terminate the string
}
Related
Why this works:
#include <stdio.h>
void slice(char *st, int m, int n)
{
int i = 0;
while ((i + m) < n)
{
st[i] = st[i + m];
i++;
}
st[i-1] = '\0';
}
int main()
{
char st[] = "Hello";
slice(st, 1, 6);
printf("The value of string is %s\n", st);
return 0;
}
And this doesn't:
#include <stdio.h>
void slice(char *st, int m, int n)
{
int i = 0;
while ((i + m) < n)
{
st[i] = st[i + m];
i++;
}
st[i-1] = '\0';
}
int main()
{
char*st = "Hello";
slice(st, 1, 6);
printf("The value of string is %s\n", st);
return 0;
}
In first I initialized my string using:
char st[]="Hello"; (using array)
And in latter I used:
char*st="Hello"; (using pointer)
I'm kind of getting confused between these 2 initialization types, what's the key difference between declaring a string by using char st[]="Hello"; and by using char*st = "Hello";.
With char st[] = "Hello";, st[] is a modifiable array of characters. The call slice(st, 1, 6); takes the array st and converts to a pointer to the first element of the array. slice() then receives that pointer, a pointer to modifiable characters.
With char *st = "Hello";, st is a pointer that points to a string literal "Hello". With the call slice(st, 1, 6);, the function receives a copy of the pointer - a pointer to the string literal. Inside slice(), code st[i] = ... is attempting to modify a string literal, that is undefined behavior (UB). It might work, it might fail, it might work today and fail tomorrow - it is not defined.
Do not attempt to modify a string literal.
... passing strings to a function ...
In both cases, code does not pass a string to slice(), but a pointer to a string. Knowing that subtle distinction helps in understanding what is truly happening.
This is an artifact of old syntax in C:
char * s = "Hello world!";
is a non-const character pointer to const memory. It is still permitted by syntax, but the string is still not a mutable object. To be pedantic it should really be written as:
const char * s = "Hello world!";
In contrast:
char s[] = "Hello world!";
allocates a local (on the stack), mutable array and copies the string data to it (from wherever the non-mutable copy is stored in memory). Your function can then do as it likes to your local copy of the string.
The type char [] is different from the type char* (char* is a variable - int. but char[] is an array which is not a variable). However, an array name can be used as a pointer to the array.
So we can say that st[] is technically similar to *str .
the problem in the 2nd version of your code
If you have read-only strings then you can use const char* st = "hello"; or simply char* st = "hello"; . So the string is most probably be stored in a read-only memory location and you'll not be able to modify it.
However, if you want to be able to modify it, use the malloc function:
char *st= (char*) malloc(n*sizeof(char)); /* n-The initial size that you need */
// ...
free(st);
**So to allocate memory for st, count the characters ("hello"-strlen(st)=5) and add 1 for this terminating null character , and functions like scanf and strcpy will add the null character;
so the code becomes :
#include <stdio.h>
void slice(char *st, int m, int n)
{
int i = 0;
while ((i + m) < n)
{
st[i] = st[i + m];
i++;
}
st[i-1] = '\0';
}
int main()
{
char *st =malloc(6*sizeof(char)) ;
const char *cpy="hello";
strcpy(st, cpy); /* copies the string pointed by cpy (including the null character) to the st. */
slice(st, 1, 6);
printf("The value of string is %s\n", st);
return 0;
}
you can fill your string also by a for loop or by scanf() .
in the case of a large allocation you must end your code with free(st);
I have a string declared as such:
char *mode_s = (char *)calloc(MODE_S_LEN, sizeof(char));
How can I add a char to the end of the array?
Lets assume " first available position " means at index 0.
char *mode_s = (char *)calloc(MODE_S_LEN, sizeof(char));
*mode_s='a';
To store a character at an arbitrary index n
*(mode_s+n)='b';
Use pointer algebra, as demonstrated above, which is equivalent to
mode_s[n]='b';
One sees that the first case simply means that n=0.
If you wish to eliminate incrementing the counter, as specified in the comment bellow, you can write a data structure and a supporting function that fits your needs. A simple one would be
typedef struct modeA{
int size;
int index;
char *mode_s;
}modeA;
The supporting function could be
int add(modeA* a, char toAdd){
if(a->size==a->index) return -1;
a->mode_s[index]=toAdd;
a->index++;
return 0;
}
It returns 0 when the add was successful, and -1 when one runs out of space.
Other functions you might need can be coded in a similar manner. Note that as C is not object oriented, the data structure has to be passed to the function as a parameter.
Finally you code code a function creating an instance
modeA genModeA(int size){
modeA tmp;
tmp.mode_s=(char *)calloc(size, sizeof(char));
tmp.size=size;
tmp.index=0;
return tmp;
}
Thus using it with no need to manually increment the counter
modeA tmp=genModeA(MODE_S_LEN);
add(&tmp,'c');
There is no standard function to concatenate a character to a string in C. You can easily define such a function:
#include <string.h>
char *strcatc(char *str, char c) {
size_t len = strlen(str);
str[len++] = c;
str[len] = '\0';
return str;
}
This function only works if str is allocated or defined with a larger size than its length + 1, ie if there is available space at its end. In your example, mode_s is allocated with a size of MODE_S_LEN, so you can put MODE_S_LEN-1 chars into it:
char *mode_s = calloc(MODE_S_LEN, sizeof(*mode_s));
for (int i = 0; i < MODE_S_LEN - 1; i++) {
strcatc(mode_s, 'X');
}
char newchar = 'a'; //or getch() from keyboard
//realloc memory:
char *mode_sNew = (char *)calloc(MODE_S_LEN + 1, sizeof(char));
//copy the str:
srncpy(mode_sNew, mode_s, MODE_S_LEN);
//put your char:
mode_sNew[MODE_S_LEN] = newchar;
//free old memory:
free(mode_s);
//reassign to the old string:
mode_s = mode_sNew;
//in a loop you can add as many characters as you want. You also can add more than one character at once, but assign only one in a new position
So I've looked around on SO and can't find code that answers my question. I have written a function that is supposed to reverse a string as input in cmd-line. Here is the function:
void reverse (char string[]) {
int x;
int i = 0;
char line[strlen(string)];
for (x = strlen(string) - 1; x > 0; x--) {
char tmp = string[x];
line[i] = tmp;
i++;
}
string = line;
}
When I call my reverse() function, the string stays the same. i.e., 'abc' remains 'abc'
If more info is needed or question is inappropriate, let me know.
Thanks!!
You're declaring your line array one char shorter remember the null at the end.
Another point, it should be for (x = strlen(string) - 1; x >= 0; x--) since you need to copy the character at 0.
void reverse (char string[]) {
int x;
int i = 0;
char line[strlen(string) + 1];
for (x = strlen(string) - 1; x >= 0; x--) {
char tmp = string[x];
line[i] = tmp;
i++;
}
for(x = 0; x < strlen(string); x++)
{
string[x] = line[x];
}
}
Note that this function will cause an apocalypse when passed an empty string or a string literal (as Bobby Sacamano said).
Suggestion you can probably do: void reverse(char source[], char[] dest) and do checks if the source string is empty.
I think that your answer is almost correct. You don't actually need an extra slot for the null character in line. You just need two minor changes:
Change the assignment statement at the bottom of the procedure to a memcpy.
Change the loop condition to <-
So, your correct code is this:
void reverse (char string[]) {
int x;
int i = 0;
char line[strlen(string)];
for (x = strlen(string) - 1; x >= 0; x--) {
char tmp = string[x];
line[i] = tmp;
i++;
}
memcpy(string, line, sizeof(char) * strlen(line));
}
Since you want to reverse a string, you first must decide whether you want to reverse a copy of the string, or reverse the string in-situ (in place). Since you asked about this in 'C' context, assume you mean to change the existing string (reverse the existing string) and make a copy of the string in the calling function if you want to preserve the original.
You will need the string library
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
Array indexing works, and this version takes that approach,
/* this first version uses array indexing */
char*
streverse_a(char string[])
{
int len; /*how big is your string*/
int ndx; /*because 'i' is hard to search for*/
char tmp; /*hold character to swap*/
if(!string) return(string); /*avoid NULL*/
if( (len=strlen(string)) < 2 ) return(string); /*one and done*/
for( ndx=0; ndx<len/2; ndx++ ) {
tmp=string[ndx];
string[ndx]=string[len-1-ndx];
string[len-1-ndx]=tmp;
}
return(string);
}
But you can do the same with pointers,
/* this is how K&R would write the function with pointers */
char*
streverse(char* sp)
{
int len, ndx; /*how big is your string */
char tmp, *bp, *ep; /*pointers to begin/end, swap temporary*/
if(!sp) return(sp); /*avoid NULL*/
if( (len=strlen(bp=sp)) < 2 ) return(sp); /*one and done*/
for( ep=bp+len-1; bp<ep; bp++, ep-- ) {
tmp=*bp; *bp=*ep; *ep=tmp; /*swap*/
}
return(sp);
}
(No, really, the compiler does not charge less for returning void.)
And because you always test your code,
char s[][100] = {
"", "A", "AB", "ABC", "ABCD", "ABCDE",
"hello, world", "goodbye, cruel world", "pwnz0r3d", "enough"
};
int
main()
{
/* suppose your string is declared as 'a' */
char a[100];
strcpy(a,"reverse string");
/*make a copy of 'a', declared the same as a[]*/
char b[100];
strcpy(b,a);
streverse_a(b);
printf("a:%s, r:%s\n",a,b);
/*duplicate 'a'*/
char *rp = strdup(a);
streverse(rp);
printf("a:%s, r:%s\n",a,rp);
free(rp);
int ndx;
for( ndx=0; ndx<10; ++ndx ) {
/*make a copy of 's', declared the same as s[]*/
char b[100];
strcpy(b,s[ndx]);
streverse_a(b);
printf("s:%s, r:%s\n",s[ndx],b);
/*duplicate 's'*/
char *rp = strdup(s[ndx]);
streverse(rp);
printf("s:%s, r:%s\n",s[ndx],rp);
free(rp);
}
}
The last line in your code does nothing
string = line;
Parameters are passed by value, so if you change their value, that is only local to the function. Pointers are the value of the address of memory they are pointing to. If you want to modify the pointer that the function was passed, you need to take a pointer to that pointer.
Here is a short example of how you could do that.
void reverse (char **string) {
char line = malloc(strlen(*string) + 1);
//automatic arrays are deallocated once the function ends
//so line needs to be dynamically or statically allocated
// do something to line
*string = line;
}
The obvious issue with this is that you can initialize the string with static memory, then this method will replace the static memory with dynamic memory, and then you'll have to free the dynamic memory. There's nothing functionally wrong with that, it's just a bit dangerous, since accidentally freeing the string literal is illegal.
char *test = "hello";
reverse(test);
free(test); //this is pretty scary
Also, if test was allocated as dynamic memory, the pointer to it would be lost and then it would become a memory leak.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int reverse(char *, int);
main()
{
char *word = "Thanks for your help";
reverse(word, strlen(word));
printf("%s", word);
getchar();
}
int reverse(char *line, int len)
{
int i, j;
char *newline = malloc(strlen(line));
for (i = len - 1, j = 0 ; i >= 0; i--, j++)
{
newline[j] = line[i];
}
newline[j] = '\0';
line = &newline;
}
Hey folks. I've got a simple C question that I can't seem to solve.
The program above is meant to take in a string and print it out backwards. Reverse is the function by which this is done.
The issue, specifically, is that when I print word in main(), the string appears unchanged. I've attempted to make the address of line the address of newline, but it doesn't have any effect.
int reverse(char *line, int len)
{
int i, j;
char *newline = malloc(strlen(line));
for (i = len - 1, j = 0 ; i >= 0; i--, j++)
{
newline[j] = line[i];
}
newline[j] = '\0';
line = &newline; // Your problem is here
}
You're merely assigning to the local line pointer. This has no effect on the calling function whatsoever.
Consider instead:
char *reverse(char *line, int len)
{
// ...
return newline;
}
Additional advice:
Turn on compiler warnings, and heed them. You've got lots of little things wrong (e.g. reverse isn't currently returning anything, but is declared as returning int).
Given that the first argument of reverse is a pointer to a C string (NUL-terminated), there's no need to take a length argument as well.
A reverse function doesn't necessarily need to be defined as returning a copy of the string, reversed. It could instead reverse a string in-place. Note that you cannot pass a string literal to a function like this, as they are read-only.
Here's how I would write this:
#include <stdio.h>
#include <string.h>
void reverse(char *str)
{
size_t i, j;
for (i = strlen(str) - 1, j = 0 ; i > j; i--, j++)
{
// Swap characters
char c = str[i];
str[i] = str[j];
str[j] = c;
}
}
int main(void)
{
// Mutable string allocated on the stack;
// we cannot just pass a string literal to reverse().
char str[] = "Here is a test string";
reverse(str);
printf("Result: \"%s\"\n", str);
return 0;
}
Note that the for loop condition is i > j, because we want each to only traverse half the array, and not swap each character twice.
Result:
$ ./a.exe
Result: "gnirts tset a si ereH"
Take a look at the code below:
void addOne(int a) {
int newA = a + 1;
a = newA;
}
int main() {
int num = 5;
addOne(num);
printf("%d\n", num);
}
Do you see why that will print 5, and not 6? It's because when you pass num to addOne, you actually make a copy of num. When addOne changes a to newA, it is changing the copy (called a), not the original variable, num. C has pass-by-value semantics.
Your code suffers from the same problem (and a couple other things). When you call reverse, a copy of word is made (not a copy of the string, but a copy of the character pointer, which points to the string). When you change line to point to your new string, newLine, you are not actually changing the passed-in pointer; you are changing the copy of the pointer.
So, how should you implement reverse? It depends: there are a couple options.
reverse could return a newly allocated string containing the original string, reversed. In this case, your function signature would be char *reverse, instead of int reverse.
reverse could modify the original string in place. That is, you never allocate a new string, and simply move the characters of the original string around. This works, in general, but not in your case because char pointers initialized with string literals do not necessarily point to writable memory.
reverse could actually change the passed-in pointer to point at a new string (what you are trying to do in your current code). To do this, you'd have to write a function void reverse(char **pointerToString). Then you could assign *pointerToString = newLine;. But this is not great practice. The original passed-in argument is now inaccessible, and if it was malloc'd, it can't be freed.
#include <stdio.h>
#include <string.h>
void reverse(char * str[]) {
int i;
int reverse = sizeof(str);
for(i=0;i<=sizeof(str);i++){
*str[i]=*str[reverse];
reverse--;
}
}
main() {
char *word;
printf("Enter a word please=>");
scanf("%s",word);
reverse(word);
printf("%s",word);
}
I am trying to get a string input and pass it to reverse() function to reverse the word entered i.e ("abcd" -> "dcba"), however i am having some difficulty using pointers.
I am not able to change the values held by the char *word in the memory.
You haven't allocated any storage for word. Change
char *word;
scanf("%s",word);
to
char word[20];
scanf("%19s",word);
There are also a number of issues with reverse
Its signature accepts an array of strings rather than a single string (an array of chars).
You need to use strlen to calculate the length of a string. sizeof tells you the size of a char** on your platform.
You reverse the string twice so will currently reverse it then reinstate the original order.
Your reverse algorithm would include reversing the nul terminator. This wouldn't normally be considered part of a string and needs to stay at the end of the array to mark the end of the string.
The following (untested) function should be closer to what you want
void reverse(char* str) {
int i;
int len = strlen(str);
for (int i=0; i<len/2; i++) {
char tmp = str[i];
str[i] = str[len-i-1];
str[len-i-1] = tmp;
}
}