I am trying to create a function that accepts a string and converts all lowercase letters to uppercase and everything else into zeros then prints the string. Here is what I have:
void upperAndZeros(char* toUpper) {
char *toReturn[(sizeof(toUpper) / sizeof(*toUpper)) + 1];
int i;
for (i = 0; toUpper[i] != '\0'; i++) {
if (toUpper[i] >= 'a' && toUpper[i] <= 'z') {
toReturn[i] = (char) toupper(toUpper[i]); //this is line 127 in the code
} else {
toReturn[i] = (char) 0;
}
}
toReturn[i] = '\0';
printf("The modified string is '%s'", toReturn);
}
But when I go to compile this I get the following error:
127:25 warning: assignment makes pointer from integer without a cast
The error you are getting is because you are trying to char in a char * array here
toReturn[i] = (char) toupper(toUpper[i]);
You want an array of char not char *. Change this line
char *toReturn[(sizeof(toUpper) / sizeof(*toUpper)) + 1];
to
char toReturn[(sizeof(toUpper) / sizeof(*toUpper)) + 1];
Another change that you should do, which was suggested in the comment is, sizeof() won't work here, as an array decays into a pointer when passed onto a function. sizeof(toUpper) would return the size of a pointer. Read this to understand the problem.
Use strlen() here, if the string pointed by toUpper is NUL terminated.
If not then send the length separately to the function as another parameter.
Lots of problems, something like this should work better:
void upperAndZeros(char* toUpper) {
char toReturn[strlen(toUpper) + 1];
int i;
for (i = 0; toUpper[i] != '\0'; i++) {
if (islower(toUpper[i])) {
toReturn[i] = (char) toupper(toUpper[i]);
} else {
toReturn[i] = '0'; // beware not 0 but char '0'
}
}
toReturn[i] = '\0';
printf("The modified string is '%s'", toReturn);
}
Related
My teacher gave us an assignment where we create a function that reads ASCII digit characters and converts them to a number without using any library functions such as atoi. Through some research i came up with this in my own file:
#include <stdio.h>
#include <sttdef.h>
int main() {
char testString[] = "123";
int convertedResult = 0;
int i;
for(i = 0; testString[i] != '\0'; i++){
convertedResult = convertedResult*10 + testString[i] - '0';
printf("%i\n",convertedResult);
if (testString[i] == '\0') {
break;
}
}
return 0;
}
While this works on its own i have to use the main file he gave us to call on this specific function.
char *asciiToInteger(char *inputString, int *integerPtr) {
return inputString;
}
I'm a bit confused as to how to proceed from here? attatched picture is main
#include <stdio.h>
#include <stddef.h>
char * asciiToInteger(char *inputString, int *integerPtr){
int convertedResult =0;
for(int i = 0; inputString[i] != '\0'; i++){
convertedResult = convertedResult*10 + inputString[i] - '0';
}
*integerPtr=convertedResult;
return inputString;
}
int main() {
char testString[] = "123";
int integerPtr;
asciiToInteger(testString, &integerPtr) ;
printf("%d\n",integerPtr);
return 0;
}
Your code has a couple of problems:
It assumes the entire string is digits
It checks for the end of string twice
I think a better implementation would be:
#include <stdio.h>
const char *asciiToInteger(const char *inputString, int *value)
{
int result = 0;
while (isdigit((unsigned int) *inputString))
{
result *= 10;
result += *inputString++ - '0';
}
*value = result;
return inputString;
}
This returns a pointer to the first non-converted character, which might be to the end of string marker if the string is all digits. I added const on the strings of course, since this converter is just reading from the strings.
When you get an assignment like this the first step is to make sure you understand what the function is supposed to do. Your question has no such description so that is the place to start.
From the behavior of the main function it seems to be something like:
If the first character in the input string is not a digit return NULL
If the first character in the input string is a digit convert all leading digits to an integer stored in the object pointed to by integerPtr and return a pointer to the character following the converted digits.
Examples:
inputString = "a123b67" --> return NULL
inputString = "123b67" --> *integerPtr = 123 and return a pointer to the 'b' in the input
That could look something like this:
char *asciiToInteger(char *inputString, int *integerPtr) {
if (*inputString < '0' || *inputString > '9')
return NULL; // no leading digit
*integerPtr = 0;
do
{
*integerPtr = *integerPtr * 10 + *inputString - '0';
++inputString;
} while (*inputString >= '0' && *inputString <= '9');
return inputString;
}
Notice that the code above can't handle negative integers.
I'm trying to print a reversed string/array. I've used the following code and it seems to be able to give my second array, revString, values in the correct order from the first array string. I am also able to print individual characters in both arrays, and I'm able to print the entire string of the first array. However the revString array doesn't print at all. I'm wondering if I am missing a huge point here.
void reverseString(char string[]) {
int i = strlen(string);
int i_2 = 0;
revString arrayet
char revString[i + 1];
char *character;
while (i > -1) {
character = &string[i];
revString[i_2] = *character;
printf("%c", revString[i_2]);
i = i - 1;
i_2 = i_2 + 1;
}
revString[i_2] = '\0';
printf("%d\n", i_2);
printf("%s", revString);
}
The code gives now the following output with example string "Hello World";
dlrow olleH13
As you can see the final printf statement doesn't do anything
In C language indexing is 0 based. so, if you make a string of 10 length, the last character will be at index 9.
In your code, when you are assigning characters to revString, your code is trying to access string[len].
your code should be like this..
int i = strlen(string) - 1;
Your code reverses the string string including the null terminator at string[i]. The resulting array starts with a null terminator, hence printf outputs nothing.
Here is a modified version:
void reverseString(char string[]) {
int i = strlen(string);
int i_2 = 0;
char revString[i + 1];
char character;
while (i > 0) {
i = i - 1;
character = string[i];
revString[i_2] = character;
//printf("%c", revString[i_2]);
i_2 = i_2 + 1;
}
revString[i_2] = '\0';
printf("%d\n", i_2);
printf("%s", revString);
}
Output:
11
dlrow olleH
For an assignment in class, we have been instructed to write a program which takes a string and a delimiter and then takes "words" and stores them in a new array of strings. i.e., the input ("my name is", " ") would return an array with elements "my" "name" "is".
Roughly, what I've attempted is to:
Use a separate helper called number_of_delimeters() to determine the size of the array of strings
Iterate through the initial array to find the number of elements in a given string which would be placed in the array
Allocate storage within my array for each string
Store the elements within the allocated memory
Include directives:
#include <stdlib.h>
#include <stdio.h>
This is the separate helper:
int number_of_delimiters (char* s, int d)
{
int numdelim = 0;
for (int i = 0; s[i] != '\0'; i++)
{
if (s[i] == d)
{
numdelim++;
}
}
return numdelim;
}
`This is the function itself:
char** split_at (char* s, char d)
{
int numdelim = number_of_delimiters(s, d);
int a = 0;
int b = 0;
char** final = (char**)malloc((numdelim+1) * sizeof(char*));
for (int i = 0; i <= numdelim; i++)
{
int sizeofj = 0;
while (s[a] != d)
{
sizeofj++;
a++;
}
final[i] = (char*)malloc(sizeofj);
a++;
int j = 0;
while (j < sizeofj)
{
final[i][j] = s[b];
j++;
b++;
}
b++;
final[i][j+1] = '\0';
}
return final;
}
To print:
void print_string_array(char* a[], unsigned int alen)
{
printf("{");
for (int i = 0; i < alen; i++)
{
if (i == alen - 1)
{
printf("%s", a[i]);
}
else
{
printf("%s ", a[i]);
}
}
printf("}");
}
int main(int argc, char *argv[])
{
print_string_array(split_at("Hi, my name is none.", ' '), 5);
return 0;
}
This currently returns {Hi, my name is none.}
After doing some research, I realized that the purpose of this function is either similar or identical to strtok. However, looking at the source code for this proved to be little help because it included concepts we have not yet used in class.
I know the question is vague, and the code rough to read, but what can you point to as immediately problematic with this approach to the problem?
The program has several problems.
while (s[a] != d) is wrong, there is no delimiter after the last word in the string.
final[i][j+1] = '\0'; is wrong, j+1 is one position too much.
The returned array is unusable, unless you know beforehand how many elements are there.
Just for explanation:
strtok will modify the array you pass in! After
char test[] = "a b c ";
for(char* t = test; strtok(t, " "); t = NULL);
test content will be:
{ 'a', 0, 'b', 0, 'c', 0, 0 }
You get subsequently these pointers to your test array: test + 0, test + 2, test + 4, NULL.
strtok remembers the pointer you pass to it internally (most likely, you saw a static variable in your source code...) so you can (and must) pass NULL the next time you call it (as long as you want to operate on the same source string).
You, in contrast, apparently want to copy the data. Fine, one can do so. But here we get a problem:
char** final = //...
return final;
void print_string_array(char* a[], unsigned int alen)
You just return the array, but you are losing length information!
How do you want to pass the length to your print function then?
char** tokens = split_at(...);
print_string_array(tokens, sizeof(tokens));
will fail, because sizeof(tokens) will always return the size of a pointer on your local system (most likely 8, possibly 4 on older hardware)!
My personal recommendation: create a null terminated array of c strings:
char** final = (char**)malloc((numdelim + 2) * sizeof(char*));
// ^ (!)
// ...
final[numdelim + 1] = NULL;
Then your print function could look like this:
void print_string_array(char* a[]) // no len parameter any more!
{
printf("{");
if(*a)
{
printf("%s", *a); // printing first element without space
for (++a; *a; ++a) // *a: checking, if current pointer is not NULL
{
printf(" %s", *a); // next elements with spaces
}
}
printf("}");
}
No problems with length any more. Actually, this is exactly the same principle C strings use themselves (the terminating null character, remember?).
Additionally, here is a problem in your own code:
while (j < sizeofj)
{
final[i][j] = s[b];
j++; // j will always point behind your string!
b++;
}
b++;
// thus, you need:
final[i][j] = '\0'; // no +1 !
For completeness (this was discovered by n.m. already, see the other answer): If there is no trailing delimiter in your source string,
while (s[a] != d)
will read beyond your input string (which is undefined behaviour and could result in your program crashing). You need to check for the terminating null character, too:
while(s[a] && s[a] != d)
Finally: how do you want to handle subsequent delimiters? Currently, you will insert empty strings into your array? Print out your strings as follows (with two delimiting symbols - I used * and + like birth and death...):
printf("*%s+", *a);
and you will see. Is this intended?
Edit 2: The variant with pointer arithmetic (only):
char** split_at (char* s, char d)
{
int numdelim = 0;
char* t = s; // need a copy
while(*t)
{
numdelim += *t == d;
++t;
}
char** final = (char**)malloc((numdelim + 2) * sizeof(char*));
char** f = final; // pointer to current position within final
t = s; // re-assign t, using s as start pointer for new strings
while(*t) // see above
{
if(*t == d) // delimiter found!
{
// can subtract pointers --
// as long as they point to the same array!!!
char* n = (char*)malloc(t - s + 1); // +1: terminating null
*f++ = n; // store in position pointer and increment it
while(s != t) // copy the string from start to current t
*n++ = *s++;
*n = 0; // terminate the new string
}
++t; // next character...
}
*f = NULL; // and finally terminate the string array
return final;
}
While I've now been shown a more elegant solution, I've found and rectified the issues in my code:
char** split_at (char* s, char d)
{
int numdelim = 0;
int x;
for (x = 0; s[x] != '\0'; x++)
{
if (s[x] == d)
{
numdelim++;
}
}
int a = 0;
int b = 0;
char** final = (char**)malloc((numdelim+1) * sizeof(char*));
for (int i = 0; i <= numdelim; i++)
{
int sizeofj = 0;
while ((s[a] != d) && (a < x))
{
sizeofj++;
a++;
}
final[i] = (char*)malloc(sizeofj);
a++;
int j = 0;
while (j < sizeofj)
{
final[i][j] = s[b];
j++;
b++;
}
final[i][j] = '\0';
b++;
}
return final;
}
I consolidated what I previously had as a helper function, and modified some points where I incorrectly incremented .
I am new to programming in C and am trying to write a simple function that will normalize a char array. At the end i want to return the length of the new char array. I am coming from java so I apologize if I'm making mistakes that seem simple. I have the following code:
/* The normalize procedure normalizes a character array of size len
according to the following rules:
1) turn all upper case letters into lower case ones
2) turn any white-space character into a space character and,
shrink any n>1 consecutive whitespace characters to exactly 1 whitespace
When the procedure returns, the character array buf contains the newly
normalized string and the return value is the new length of the normalized string.
*/
int
normalize(unsigned char *buf, /* The character array contains the string to be normalized*/
int len /* the size of the original character array */)
{
/* use a for loop to cycle through each character and the built in c functions to analyze it */
int i;
if(isspace(buf[0])){
buf[0] = "";
}
if(isspace(buf[len-1])){
buf[len-1] = "";
}
for(i = 0;i < len;i++){
if(isupper(buf[i])) {
buf[i]=tolower(buf[i]);
}
if(isspace(buf[i])) {
buf[i]=" ";
}
if(isspace(buf[i]) && isspace(buf[i+1])){
buf[i]="";
}
}
return strlen(*buf);
}
How can I return the length of the char array at the end? Also does my procedure properly do what I want it to?
EDIT: I have made some corrections to my program based on the comments. Is it correct now?
/* The normalize procedure normalizes a character array of size len
according to the following rules:
1) turn all upper case letters into lower case ones
2) turn any white-space character into a space character and,
shrink any n>1 consecutive whitespace characters to exactly 1 whitespace
When the procedure returns, the character array buf contains the newly
normalized string and the return value is the new length of the normalized string.
*/
int
normalize(unsigned char *buf, /* The character array contains the string to be normalized*/
int len /* the size of the original character array */)
{
/* use a for loop to cycle through each character and the built in c funstions to analyze it */
int i = 0;
int j = 0;
if(isspace(buf[0])){
//buf[0] = "";
i++;
}
if(isspace(buf[len-1])){
//buf[len-1] = "";
i++;
}
for(i;i < len;i++){
if(isupper(buf[i])) {
buf[j]=tolower(buf[i]);
j++;
}
if(isspace(buf[i])) {
buf[j]=' ';
j++;
}
if(isspace(buf[i]) && isspace(buf[i+1])){
//buf[i]="";
i++;
}
}
return strlen(buf);
}
The canonical way of doing something like this is to use two indices, one for reading, and one for writing. Like this:
int normalizeString(char* buf, int len) {
int readPosition, writePosition;
bool hadWhitespace = false;
for(readPosition = writePosition = 0; readPosition < len; readPosition++) {
if(isspace(buf[readPosition]) {
if(!hadWhitespace) buf[writePosition++] = ' ';
hadWhitespace = true;
} else if(...) {
...
}
}
return writePosition;
}
Warning: This handles the string according to the given length only. While using a buffer + length has the advantage of being able to handle any data, this is not the way C strings work. C-strings are terminated by a null byte at their end, and it is your job to ensure that the null byte is at the right position. The code you gave does not handle the null byte, nor does the buffer + length version I gave above. A correct C implementation of such a normalization function would look like this:
int normalizeString(char* string) { //No length is passed, it is implicit in the null byte.
char* in = string, *out = string;
bool hadWhitespace = false;
for(; *in; in++) { //loop until the zero byte is encountered
if(isspace(*in) {
if(!hadWhitespace) *out++ = ' ';
hadWhitespace = true;
} else if(...) {
...
}
}
*out = 0; //add a new zero byte
return out - string; //use pointer arithmetic to retrieve the new length
}
In this code I replaced the indices by pointers simply because it was convenient to do so. This is simply a matter of style preference, I could have written the same thing with explicit indices. (And my style preference is not for pointer iterations, but for concise code.)
if(isspace(buf[i])) {
buf[i]=" ";
}
This should be buf[i] = ' ', not buf[i] = " ". You can't assign a string to a character.
if(isspace(buf[i]) && isspace(buf[i+1])){
buf[i]="";
}
This has two problems. One is that you're not checking whether i < len - 1, so buf[i + 1] could be off the end of the string. The other is that buf[i] = "" won't do what you want at all. To remove a character from a string, you need to use memmove to move the remaining contents of the string to the left.
return strlen(*buf);
This would be return strlen(buf). *buf is a character, not a string.
The notations like:
buf[i]=" ";
buf[i]="";
do not do what you think/expect. You will probably need to create two indexes to step through the array — one for the current read position and one for the current write position, initially both zero. When you want to delete a character, you don't increment the write position.
Warning: untested code.
int i, j;
for (i = 0, j = 0; i < len; i++)
{
if (isupper(buf[i]))
buf[j++] = tolower(buf[i]);
else if (isspace(buf[i])
{
buf[j++] = ' ';
while (i+1 < len && isspace(buf[i+1]))
i++;
}
else
buf[j++] = buf[i];
}
buf[j] = '\0'; // Null terminate
You replace the arbitrary white space with a plain space using:
buf[i] = ' ';
You return:
return strlen(buf);
or, with the code above:
return j;
Several mistakes in your code:
You cannot assign buf[i] with a string, such as "" or " ", because the type of buf[i] is char and the type of a string is char*.
You are reading from buf and writing into buf using index i. This poses a problem, as you want to eliminate consecutive white-spaces. So you should use one index for reading and another index for writing.
In C/C++, a native string is an array of characters that ends with 0. So in essence, you can simply iterate buf until you read 0 (you don't need to use the len variable at all). In addition, since you are "truncating" the input string, you should set the new last character to 0.
Here is one optional solution for the problem at hand:
int normalize(char* buf)
{
char c;
int i = 0;
int j = 0;
while (buf[i] != 0)
{
c = buf[i++];
if (isspace(c))
{
j++;
while (isspace(c))
c = buf[i++];
}
if (isupper(c))
buf[j] = tolower(c);
j++;
}
buf[j] = 0;
return j;
}
you should write:
return strlen(buf)
instead of:
return strlen(*buf)
The reason:
buf is of type char* - it's an address of a char somewhere in the memory (the one in the beginning of the string). The string is null terminated (or at least should be), and therefore the function strlen knows when to stop counting chars.
*buf will de-reference the pointer, resulting on a char - not what strlen expects.
Not much different then others but assumes this is an array of unsigned char and not a C string.
tolower() does not itself need the isupper() test.
int normalize(unsigned char *buf, int len) {
int i = 0;
int j = 0;
int previous_is_space = 0;
while (i < len) {
if (isspace(buf[i])) {
if (!previous_is_space) {
buf[j++] = ' ';
}
previous_is_space = 1;
} else {
buf[j++] = tolower(buf[i]);
previous_is_space = 0;
}
i++;
}
return j;
}
#OP:
Per the posted code it implies leading and trailing spaces should either be shrunk to 1 char or eliminate all leading and trailing spaces.
The above answer simple shrinks leading and trailing spaces to 1 ' '.
To eliminate trailing and leading spaces:
int i = 0;
int j = 0;
while (len > 0 && isspace(buf[len-1])) len--;
while (i < len && isspace(buf[i])) i++;
int previous_is_space = 0;
while (i < len) { ...
I am new to cpp and have a question regarding arrays. The code I have below should create a reversed version of str and have it be stored in newStr. However, newStr always comes up empty. Can someone explain to me why this is happening even though I am assigning a value from str into it?
void reverse (char* str) {
char* newStr = (char*)malloc(sizeof(str));
for (int i=0;i<sizeof(str)/sizeof(char);i++) {
int index = sizeof(str)/sizeof(char)-1-i;
newStr [i] = str [index];
}
}
PS: I know that it is much more efficient to reverse an array by moving the pointer or by using the std::reverse function but I am interested in why the above code does not work.
As above commenters pointed out sizeof(str) does not tell you the length of the string. You should use size_t len = strlen(str);
void reverse (char* str) {
size_t len = strlen(str);
char* newStr = (char*)malloc(len + 1);
for (int i=0; i<len;i++) {
int index = len-1-i;
newStr[i] = str[index];
}
newStr[len] = '\0'; // Add terminator to the new string.
}
Don't forget to free any memory you malloc. I assume your function is going to return your new string?
Edit: +1 on the length to make room for the terminator.
The sizeof operator (it is not a function!) is evaluated at compile time. You are passing it a pointer to a region of memory that you claim holds a string. However, the length of this string isn't fixed at compile time. sizeof(str)/sizeof(char) will always yield the size of a pointer on your architecture, probably 8 or 4.
What you want is to use strlen to determine the length of your string.
Alternatively, a more idiomatic way of doing this would be to use std::string (if you insist of reversing the string yourself)
std::string reverse(std::string str) {
for (std::string::size_type i = 0, j = str.size(); i+1 < j--; ++i) {
char const swap = str[i];
str[i] = str[j];
str[j] = swap;
}
return str;
}
Note that due to implicit conversion (see overload (5)), you can also call this function with your plain C-style char pointer.
There are two issues here:
The sizeof operator won't give you the length of the string. Rather, it gives you the size of a char* on the machine you are using. You can use strlen() instead to get the
A c-string is terminated by a NULL character (which is why strlen() can return the correct length of the string). You need to make sure you are not accidentally copying the NULL character from your source string to the beginning of your destination string. Also, you need to add a NULL character at the end of your destination string or you will get some unexpected output.
#include <bits/stdc++.h>
using namespace std;
vector<string> split_string(string);
// Complete the reverseArray function below.
vector<int> reverseArray(vector<int> a) {
return {a.rbegin(), a.rend()};
}
int main()
{
ofstream fout(getenv("OUTPUT_PATH"));
int arr_count;
cin >> arr_count;
cin.ignore(numeric_limits<streamsize>::max(), '\n');
string arr_temp_temp;
getline(cin, arr_temp_temp);
vector<string> arr_temp = split_string(arr_temp_temp);
vector<int> arr(arr_count);
for (int i = 0; i < arr_count; i++) {
int arr_item = stoi(arr_temp[i]);
arr[i] = arr_item;
}
vector<int> res = reverseArray(arr);
for (int i = 0; i < res.size(); i++) {
fout << res[i];
if (i != res.size() - 1) {
fout << " ";
}
}
fout << "\n";
fout.close();
return 0;
}
vector<string> split_string(string input_string) {
string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) {
return x == y and x == ' ';
});
input_string.erase(new_end, input_string.end());
while (input_string[input_string.length() - 1] == ' ') {
input_string.pop_back();
}
vector<string> splits;
char delimiter = ' ';
size_t i = 0;
size_t pos = input_string.find(delimiter);
while (pos != string::npos) {
splits.push_back(input_string.substr(i, pos - i));
i = pos + 1;
pos = input_string.find(delimiter, i);
}
splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));
return splits;
}