Is this possible? That is, append a single char to a char*?
char temp[10] = "";
while (isalnum(*x)) {
strcat(temp, *x);
x++;
}
enqueue(&head,temp);
I have a queue which has a char* as a variable. Then I have a char* to be read, and if the char that it's pointing to is alphanumeric, I want it to append to temp. And after that, enqueue it.
Since x appears to be a char *, you can use strncat().
while (isalnum(*x)) {
if (bytes_left_in_temp > 0) {
strncat(temp, x, 1);
}
}
More efficiently, you can count how items in x can be copied, and then pass the count instead of 1.
int k = 0;
while (isalnum(x[k])) {
if (k >= bytes_left_in_temp) break;
++k;
}
strncat(temp, x, k);
However, you should make sure you take care not to append past what temp can hold, as illustrated above.
Assuming temp will allways be large enough:
char temp[10] = "";
char *p = temp;
while (isalnum(*x)) {
*p++ = *x++;
}
*p = '\0';
enqueue(&head,temp);
If you're not sure if temp will allways be large enough, add some checks (which shouldn't need
detailed explanation).
If x is char*, then you can use strcat(temp,x)
Related
I am trying to reverse a string (character array) using the following code, but when I attempt to print the string, the value of null. This is a homework assignment, but I am trying to learn so any help would be appreciated.
void input_reverse_string(const char* inputStr, char* reverseStr)
{
int i = 0;
int length = 0;
for (; *(inputStr++) != '\0'; i++)
{
length++;
}
while (*inputStr)
{
*reverseStr = *inputStr;
inputStr++;
reverseStr++;
}
const char* chr_ptr = &inputStr[length - 1];
printf("I see a %s\n", *chr_ptr);
*reverseStr = '\0';
printf("%d", length);
/* return reverseStr; */
}
Several things are out of order:
That's a strange way of computing the length of a string. You are using an index variable that you don't need, and incrementing 3 things at the same time, it's unneeded to say the least.
After calculating the length, and incrementing the inputStr pointer up to its end, you don't reset the pointer, so it still points to the end of the string (actually, one after the end!).
Inside the while you are advancing both pointers (inputStr and reverseStr) in the same direction, which can't possibly be right if you want to reverse the string.
The correct way to do this would be:
Compute the length of the string. Either use strlen() or do it by hand, but you really only need to increment one variable to do this. You can avoid incrementing inputStr, just use a temporary pointer.
Start from inputStr + length and walk backwards. Either use a pointer and do -- or just index the string).
Here's a working example:
void reverse_string(const char* inputStr, char* reverseStr) {
unsigned len = 0;
int i;
while (inputStr[len])
len++;
for (i = len - 1; i >= 0; i--) {
reverseStr[len - i - 1] = inputStr[i];
}
reverseStr[len] = '\0';
}
int main(void) {
char a[6] = "hello";
char b[6];
reverse_string(a, b);
puts(b);
return 0;
}
Output:
olleh
I am working to learn C using Kochan's Programming in C 4th edition. problem 9.7 the goal is to insert a string of characters into another array. I am supposed to write a function to accomplish this. I have two problems.
When I have the algorithm print the result as it goes through the if statements, it produces the desired output, however when I change it to an %s, I only get a partial output. My hunch is that a null character is being placed where i do not want it, but I simply cannot see it.
To see what was happening, I added a printf that would track the letter and the array space it was occupying. I was surprised to see that the first letter was not 0, but was blank, and the next letter was assigned the 0. Any insight into this would be appreciated.
The funtion of interest is "insertString".
#include <stdio.h>
#include <stdbool.h>
char x[] = {"the wrong son was shot that day"};
char text[] = {"per"};
int countString (char x[])
{
int counter, z;
for (counter = 0; x[counter] != '\0'; ++counter)
z = counter+1;
return z;
}
void insertString (char text[],char x[],int n) //source, text to input, where
{
int count, clock, i = countString(text), q = countString(x);
int counter = 0;
char y[i + q];
for(count = 0; x[count] != '\0'; ++count){
if (count < n){
y[count] = x[count];
printf("%c %i", y[count], count); //The integer call is just to put a number next to the
//letter. This is where my second issue is shown.
}
else if (counter <= i){
y[count] = text[counter];
++counter;
printf("%c", y[count]);
}
else{
y[count]= x[count - counter];
printf("%c", y[count]);
}
}
printf("\n\n");
y[count-counter] = '\0';
printf("%s", y);
}
int main (void)
{
void insertString(char text[], char x[], int i);
int countString(char x[]);
int i;
insertString(text, x, 10);
return 0;
}
10 out of 10 times I post here it is because im doing something dumb, so I use SO as an absolute last resort if i am getting into the territory of just randomly trying stuff with no methodology. Thanks for your patience in advance.
Your condition is wrong in the for. It should be x[count - counter] != '\0'
In the second condition use just < to avoid overindexing. (else if (counter < i))
You put the terminating NULL char at wrong place. You should do this: y[count] = '\0'
printf inside a string routine like this is fine for debugging, but it's a poor way to write a general-purpose function because it makes it impossible to use its output for further programmatic manipulation. It can also make it difficult to reason about how the state of the function interacts in unpredictable ways with the state of the printed data.
I assume you haven't learned about dynamic memory allocation which is a prerequisite to returning strings from functions. You can inline the function logic into main or printf only at the end of the function in the meantime.
Adding to this point, a void function would need to reallocate space in the string to insert into and would be in-place. This seems likely less generally useful than allocating a new string to hold the result.
Using global variables like char x[] when there's no need is poor practice. It's better to put those strings scoped to main. Since your function can access these variables in addition to its parameters, confusion can ensue when scope and encapsulation is breached.
Use consistent formatting and avoid variable names like q that mean virtually nothing. Instead of adding comments to explain poor var names:
void insertString (char text[],char x[],int n) //source, text to input, where
You can simply name the variables exactly what they represent:
void insertString(char *dest, char *source, int add_index)
Also, now that you've mastered countString, you can abstract this by calling the builtin strlen.
Be sure to allocate enough space in buffers: char y[i + q]; should be y[i+q+1] to allow room for the null terminator '\0'.
As for the logic, I think it's easier to break into three loops without conditions instead of one loop with conditions. This makes it easier to break the problem down into the three constituent steps:
Add everything up until add_index from the dest string to the result.
Add everything in the source string to the result.
Add everything after add_index from the dest string to the result.
Using this approach, all that's left is figuring out how to map the indexes appropriately. Here it is in code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *insert_string(char *dest, char *source, int add_index) {
int source_len = strlen(source);
int dest_len = strlen(dest);
int result_size = source_len + dest_len + 1;
char *result = malloc(result_size);
for (int i = 0; i < add_index; i++) {
result[i] = dest[i];
}
for (int i = 0; i < source_len; i++) {
result[i+add_index] = source[i];
}
for (int i = add_index; i < dest_len; i++) {
result[i+add_index] = dest[i];
}
result[result_size-1] = '\0';
return result;
}
int main(void) {
char *result = insert_string("hello world", "cruel ", 6);
printf("%s\n", result);
free(result);
return 0;
}
Although this is likely for instructional purposes, these operations can be abstracted further using builtin string functions like strncpy and sprintf.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *insert_string(char *dest, char *source, int add_index) {
int result_size = strlen(dest) + strlen(source) + 1;
char *result = malloc(result_size);
char pre[add_index+1];
pre[add_index] = '\0';
strncpy(pre, dest, add_index);
sprintf(result, "%s%s%s", pre, source, dest + add_index);
return result;
}
int main(void) {
char *result = insert_string("hello world", "cruel ", 6);
printf("%s\n", result);
free(result);
return 0;
}
Doing this in-place is more straightforward. Since the result already has the prefix, you can copy the destination postfix to create a source-sized gap in the middle and then overwrite the gap using the source string. It's up to the caller to make sure that the destination buffer is large enough to hold the insertion.
#include <stdio.h>
#include <string.h>
void insert_string(char *dest, char *source, int add_index) {
int source_len = strlen(source);
int dest_len = strlen(dest);
for (int i = add_index; i < dest_len; i++) {
dest[i+add_index] = dest[i];
}
for (int i = 0; i < source_len; i++) {
dest[i+add_index] = source[i];
}
}
int main(void) {
// allocate extra space in the string to hold the insertion
char greeting[32] = "hello world";
insert_string(greeting, "cruel ", 6);
printf("%s\n", greeting);
return 0;
}
A note of caution: none of these functions handle errors at all, so they're unsafe. Correct functions should check that the add_index falls within the bounds of the dest string. This is an exercise for the reader.
The original exercise is here:
Your function is not doing it. You need to insert the string into another string not to create a new one with both mixed. You can do it this way of course and then copy it into the original one - but it is the most uneficient way to archive it (memory & timewise).
Use the correct types.
size_t mystrlen(const char *str)
{
const char *end = str;
while(*end++);
return end - str - 1;
}
char *strinsert(char *dest, size_t pos, const char *istr)
{
char *temp = dest, *work;
size_t ilen = mystrlen(istr);
size_t nmove;
while(*temp) temp++;
nmove = temp - dest - pos + 1;
work = temp;
temp += ilen;
while(nmove--) *temp-- = *work--;
work = dest + pos;
while(*istr) *work++ = *istr++;
return dest;
}
int main()
{
char dest[128] = "0123456789012345678901234567890123456789";
printf("%s", strinsert(dest, 7, "ABCD"));
}
https://godbolt.org/z/KMnLU2
I have a class that is meant to return a char** by splitting one char* into sentences. I can allocate the memory and give it values at a certain point, but by the time I try to return it, it's completely missing.
char **makeSentences(char *chapter, int *nSentences){
int num = *nSentences;
char* chap = chapter;
char **sentences;
sentences = (char**) malloc(sizeof(char*) * num);
int stops[num + 1];
stops[0] = 0;
int counter = 0;
int stop = 1;
while (chap[counter] != '\0'){
if (chap[counter] == '.'){
stops[stop] = counter + 1;
printf("Place: %d\nStop Number: %d\n\n", counter, stop);
stop++;
}
counter++;
}
for (int i = 0; i < num; i++){
int length = stops[i+1] - stops[i];
char characters[length+1];
memcpy(characters, &chap[stops[i]], length);
characters[length] = '\0';
char *sentence = characters;
sentences[i] = sentence;
printf("%s\n",sentence);
printf("%s\n", sentences[i]);
}
char* testChar = sentences[0];
printf("%s\n", sentences[0]);
printf("%s]n", testChar);
return sentences;
}
The last two printing lines don't print anything but a newline, while the exact same lines (in the for loop) print as expected. What is going on here?
The problem is these three lines:
char characters[length+1];
char *sentence = characters;
sentences[i] = sentence;
Here you save a pointer to a local variable. That variable characters will go out of scope every iteration of the loop, leaving you with an "array" of stray pointers.
While not standard in C, almost all systems have a strdup function whichg duplicates a string by calling malloc and strcpy. I suggest you use it (or implement your own).
My head is getting bad trying to find a solution for this assignment because my ideas didn't work...
I have to interlace two char strings using pointers. See the following example (this example is not code):
char s1 = "My House Black"
char s2 = "Are very near"
Result: "MAyr eH ovuesrey Bnleaacrk"
How can I do this?
Try:
int total = strlen(char1) + strlen(char2);
int i1 = 0, i2 = 0;
for(i = 0; i < total; i++)
{
if(i % 2 == 0)
{
result[i] = char1[i1];
i1++;
}
else
{
result[i] = char2[i2];
i2++;
}
}
Here's a hint (pseudocode):
result = ""
for i = 0 to longest string's length:
result += some character (whose?)
result += another character (also, whose?)
Be careful: you need a little check somewhere, otherwise bad things might happen.
Since this is homework, I will just give you an outline.
First of all declare your two strings:
const char *s1 = "My House Black";
const char *s2 = "Are very near";
Next declare two pointers to char:
char *p1 = s1;
char *p2 = s2;
Now enter a while loop. The condition should be that *p1 or *p2 are not equal to zero.
Inside the loop output *p1 if it is not zero and then output *p2 if it is not zero. Increment each pointer if it refers to a non-zero character.
That's it, you are done!
Since this is tagged homework, I don't want to directly post code. But create three character arrays, one for each input, one long enough to contain the output, and traverse the input character arrays one character at a time (use pointer arithmetic). Store the character into your output string. Continue until you reach the end of each string. Don't forget the null terminations!
You need a target string that is large enough to hold both input strings and the string terminator.
Then you should probably use a loop (while or for) where you copy one character from each input string in each iteration.
For extra credits:
Consider the case where the input strings are of unequal length.
Is this what you want?
char * s1 = "My House Black";
char * s2 = "Are very near";
char * s = (char *)malloc(strlen(s1) + strlen(s2) + 1);
char * p1 = s1;
char * p2 = s2;
char * p = s;
while (*p1 && *p2)
{
*p++ = *p1++;
*p++ = *p2++;
}
while (*p1)
{
*p++ = *p1++;
}
while (*p2)
{
*p++ = *p2++;
}
*p = '\0';
I'm starting to learn C by reading K&R and going through some of the exercises. After some struggling, I was finally able to complete exercise 1-19 with the code below:
/* reverse: reverse the character string s */
void reverse(char s[], int slen)
{
char tmp[slen];
int i, j;
i = 0;
j = slen - 2; /* skip '\0' and \n */
tmp[i] = s[j];
while (i <= slen) {
++i;
--j;
tmp[i] = s[j];
}
/* code from copy function p 29 */
i = 0;
while ((s[i] = tmp[i]) != '\0')
++i;
}
My question is regarding that last bit of code where the tmp char array is copied to s. Why doesn't a simple s = tmp; work instead? Why does one have to iterate through the array copying index by index?
Maybe I'm just old and grumpy, but the other answers I've seen seem to miss the point completely.
C does not do array assignments, period. You cannot assign one array to another array by a simple assignment, unlike some other languages (PL/1, for instance; Pascal and many of its descendants too - Ada, Modula, Oberon, etc.). Nor does C really have a string type. It only has arrays of characters, and you can't copy arrays of characters (any more than you can copy arrays of any other type) without using a loop or a function call. [String literals don't really count as a string type.]
The only time arrays are copied is when the array is embedded in a structure and you do a structure assignment.
In my copy of K&R 2nd Edition, exercise 1-19 asks for a function reverse(s); in my copy of K&R 1st Edition, it was exercise 1-17 instead of 1-19, but the same question was asked.
Since pointers have not been covered at this stage, the solution should use indexes instead of pointers. I believe that leads to:
#include <string.h>
void reverse(char s[])
{
int i = 0;
int j = strlen(s) - 1;
while (i < j)
{
char c = s[i];
s[i++] = s[j];
s[j--] = c;
}
}
#ifdef TEST
#include <stdio.h>
int main(void)
{
char buffer[256];
while (fgets(buffer, sizeof(buffer), stdin) != 0)
{
int len = strlen(buffer);
if (len == 0)
break;
buffer[len-1] = '\0'; /* Zap newline */
printf("In: <<%s>>\n", buffer);
reverse(buffer);
printf("Out: <<%s>>\n", buffer);
}
return(0);
}
#endif /* TEST */
Compile this with -DTEST to include the test program and without to have just the function reverse() defined.
With the function signature given in the question, you avoid calling strlen() twice per line of input. Note the use of fgets() — even in test programs, it is a bad idea to use gets(). The downside of fgets() compared to gets() is that fgets() does not remove the trailing newline where gets() does. The upsides of fgets() are that you don't get array overflows and you can tell whether the program found a newline or whether it ran out of space (or data) before encountering a newline.
Your tmp array was declared on stack and so when your method completes, the memory used to hold the values will be freed because of scoping.
s = tmp means that s should point to the same memory location as tmp. This means that when tmp is freed, s will still be pointing to a now possible invalid, freed memory location.
This type of error is referred to as a dangling pointer.
Edit: This isn't a dangling modifier as pointed out in the comments of this answer. The issue is that saying s = tmp only changes what the parameter points to, not what the actual array that was passed.
Also, you could perform your reverse with a single pass and without allocating a whole array in memory by just swapping the values in place one by one:
void reverse(char s[], int slen) {
int i = 0; // First char
int j = slen - 2; // Last char minus \n\0
char tmp = 0; // Temp for the value being swapped
// Iterate over the array from the start until the two indexes collide.
while(i < j) {
tmp = s[i]; // Save the eariler char
s[i] = s[j]; // Replace it with the later char
s[j] = tmp; // Place the earlier char in the later char's spot
i++; // Move forwards with the early char
j--; // Move backwards with the later char
}
}
Because both s and tmp are memory addressees. If you s = tmp, both pointers would point to the same array.
Suppose that we have
char s[] ="ab";
/*
* Only for explanatory purposes.
*
*/
void foo(char s[]){
char tmp [] = "cd";
s= tmp;
}
foo(s);
after s= tmp you would have
s[0] : 'c'
s[1] : 'd'
s[2] : '\0'
Even though both arrays have the same data, a change in tmp, will affect both of them, because both arrays are actually the same. They both contain data that´s in the same memory address. So by changing any position of the tmp array, or destroying the tmp array, s would be affected in the same way.
By looping over the array, what you are doing is moving a piece of data from one memory address to another.
In my copy of K & R, pointers are explained in chapter 4. A quick glance through the first pages may be of help.
To round out the discussion here are two other possible ways to reverse as string:
void reverse(char string1[], char string2[])
{
int i = 0, len = 0;
while(string2[len] != '\0') // get the length of the string
len++;
while(len > 0)
{
string1[i] = string2[len-1]; // copy the elements in reverse
i++;
len--;
}
string1[i] = '\0'; // terminate the copied string
}
Or recursively:
void reverse (const char *const sPtr)
{
//if end of string
if (sPtr[0] == '\0')
{
return;
}
else //not end of the string...
{
reverse(&sPtr[1]); //recursive step
putchar(sPtr[0]); //display character
}
}
because tmp is a pointer, and you need to get a copy, not a "link".
In case of s=tmp, the value of tmp which is the also the beginning address of the array, would get copied to s.
That way both s and tmp will point to the same address in memory, which I think is not the purpose.
cheers
Try experimenting and see what happens when you do things like this:
void modifyArrayValues(char x[], int len)
{
for (int i = 0; i < len; ++i)
x[i] = i;
}
void attemptModifyArray(char x[], int len)
{
char y[10];
for (int i = 0; i < len; ++i)
y[i] = i;
x = y;
}
int main()
{
int i = 0;
char x[10];
for (i = 0; i < 10; ++i)
x[i] = 0;
attemptModifyArray(x, 10);
for (i=0; i < 10; ++i)
printf("%d\n", x[i]); // x is still all 0's
modifyArrayValues(x, 10);
for (i=0; i < 10; ++i)
printf("%d\n", x[i]); // now x has 0-9 in it
}
What happens when you modify the array directly in attemptModifyArray, you are just overwriting a local copy of the address of the array x. When you return, the original address is still in main's copy of x.
When you modify the values in the array in modifyArrayValues, you are modifying the actual array itself which has its address stored in modifyArrayValues local copy of x. When you return, x is still holding on to the same array, but you have modified the values in that array.
There's an interesting sub-thread in this thread about arrays and pointers
I found this link on wikipedia with a peculiar code snippet showing just how 'plasticine' C can be!
/* x designates an array */
x[i] = 1;
*(x + i) = 1;
*(i + x) = 1;
i[x] = 1; /* strange, but correct: i[x] is equivalent to *(i + x) */
Of course what's even more confusing in C is that I can do this:
unsigned int someval = 0xDEADD00D;
char *p = (char *)&someval;
p[2] = (char)0xF0;
So the interchangibility of pointers and arrays seems so deep-set in the C language as to be almost intentional.
What does everyone else think?
---Original Post---
s and tmp are both pointers so doing s = tmp will simply make s point at the address where tmp lives in memory.
Another problem with what you outlined is that tmp is a local variable so will become 'undefined' when it goes out of scope i.e when the function returns.
Make sure you thoroughly grasp these three concepts and you won't go far wrong
Scope
The difference between the stack and the heap
Pointers
Hope that helps and keep going!
A very straight forward answer would be -
both s and tmp are pointers to a memory location and not the arrays themselves.
In other words, s and tmp are memory addresses where the array values are stored but not the values themselves.
And one of the common ways to access these array values are by using indices like s[0] or tmp[0].
Now, if you will try to simply copy, s = tmp, the memory address of tmp array will be copied over to s. This means that, the original s array will be lost and even s memory pointer will now point to tmp array.
You will understand these concepts well with due time so keep going through the book.
I hope this elementary explanation helps.