char array printing 1 char short of what I assigned - c

I am trying to create a character array with X number of characters.
I need the first X-1 characters to be spaces and I need the Xth character to be an *.
I have written the following:
int i = 0;
int X = 5;
char spaces[X]; //In this case X is 5 so the array should have indexes 0 - 4
for(i = 0; i < X; i++) {
spaces[i] = '*'; //I start by setting all 5 char's equal to '*'
printf("spaces = '%s'\n", spaces); //This was to make sure it ran the correct # of times
}
The output of this segment is the following, the 'gh' is different every time:
spaces = '*gh'
spaces = '**h'
spaces = '***'
spaces = '****'
spaces = '****'
why does spaces only grow to 4 instead of 5 characters?
Shouldn't spaces[4] = '*'; have been called?
After setting the whole string equal to '*' I run a second for loop:
for(i = 0; i < X-1; i++) {
spaces[i] = ' ';
}
which should then set everything but the Xth character equal to ' ', but since the string is acting like its only X-1 characters long, the whole thing is set to spaces and it comes out like this
spaces = ' ';
4 spaces, when I need 4 spaces followed by an *.

You are missing the string termination character \0, which is needed once you want to print your array as a string using printf("%s",...).
So make your array one item larger than the items you want to print, and initialize it with 0, such that everything you write into the array will at the end be a valid string. Otherwise you yield undefined behaviour:
int main (void)
{
#define X 5
int i = 0;
char spaces[X+1] = { 0 };
for(i = 0; i < X; i++) {
spaces[i] = '*';
printf("spaces = '%s'\n", spaces);
}
}

In order to set first X-1 characters to be spaces and the Xth character to be an .
This will always have the last character a ''
for(i = 0; i < X-1; i++) {
spaces[i] = ' ';
spaces[i+1] = '*';
printf("spaces = '%s'\n", spaces);
}

Related

Print array of strings vertically

I know how to print a single string vertically.
char test[100] = "test";
int i;
for(i=0;i<strlen(test);i++){
printf("%c\n",test[i]);
}
Which will give me:
t
e
s
t
But how can I print an array of strings vertically? For example:
char listOfTest[2][10] = {"testing1","quizzing"};
So it can return:
tq
eu
si
tz
iz
gi
1g
Simply loop through the first string and print each character at index i in the first string and the second string till you reach the null terminator of first string
NOTE: this only work when string 1 and string2 are equal in length and will need modification for other test cases
#include <stdio.h>
int main(void)
{
char listOfTest[2][10] = {"testing1","quizzing"};
int i = 0;
//loop through string 1 till NULL is reach
while (listOfTest[0][i])
{
//prints char at index i in string 1 and 2
printf("%c%c\n", listOfTest[0][i], listOfTest[1][i]);
//increment the index value
i++;
}
return (0);
}
Simply print a character from both strings.
(Better to test for the null character rather than repeatedly call strlen().)
for(i = 0; listOfTest[0][i] && listOfTest[1][i]; i++) {
printf("%c%c\n", listOfTest[0][i], listOfTest[1][i]);
}
To extend to n strings ...
size_t num_of_strings = sizeof listOfTest/sizeof listOfTest[0];
bool done = false;
for (size_t i = 0; listOfTest[0][i] && !done; i++) {
for (size_t n = 0; n < num_of_strings; n++) {
if (listOfTest[n][i] == '\0') {
done = true;
break;
}
printf("%c", listOfTest[n][i]);
}
printf("\n");
}

c string: put ' ' if a word found in the sentence

I made a code and my target is to put spacewhere the input word was found in a sentence.
i neet to replece the small word with space
like:
Three witches watched three watches
tch
output:
Three wi es wa ed three wa es
I made this code:
#include<stdio.h>
#define S 8
#define B 50
void main() {
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
for (i = 0; i < B; i++)
{
for(j=0;j<S;j++)
{
if(small[j]!=big[i])
{
j=0;
break;
}
if(small[j]=='\0')
{
while (i-(j-1)!=i)
{
i = i - j;
big[i] = '\n';
i++;
}
}
}
}
puts(big);
}
First of all, in your exemple you work with newline '\n' and not with space.
Consider this simple example:
#include<stdio.h>
#define S 8
#define B 50
void main() {
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int cpt = 0;
int smallSize = 0;
// loop to retrieve smallSize
for (i = 0; i < S; i++)
{
if (small[i] != '\0')
smallSize++;
}
// main loop
for (i = 0; i < B; i++)
{
// stop if we hit the end of the string
if (big[i] == '\0')
break;
// increment the cpt and small index while the content of big and small are equal
if (big[i] == small[j])
{
cpt++;
j++;
}
// we didn't found the full small word
else
{
j = 0;
cpt = 0;
}
// test if we found the full word, if yes replace char in big by space
if (cpt == smallSize)
{
for (int k = 0; k < smallSize; k++)
{
big[i-k] = ' ';
}
j = 0;
cpt = 0;
}
}
puts(big);
}
You need first to retrieve the real size of the small array.
Once done, next step is to look inside "big" if there is the word small inside. If we find it, then replace all those char by spaces.
If you want to replace the whole small word with a single space, then you'll need to adapt this example !
I hope this help !
A possible way is to use to pointers to the string, one for reading and one for writing. This will allow to replace an arbitrary number of chars (the ones from small) with a single space. And you do not really want to nest loops but une only one to process every char from big.
Last but not least, void main() should never be used except in stand alone environment (kernel or embedded development). Code could become:
#include <stdio.h>
#define S 8
#define B 50
int main() { // void main is deprecated...
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int k = 0; // pointer to written back big
for (i = 0; i < B; i++)
{
if (big[i] == 0) break; // do not process beyond end of string
if(small[j]!=big[i])
{
for(int l=0; l<j; l++) big[k++] = small[l]; // copy an eventual partial small
big[k++] = big[i]; // copy the incoming character
j=0; // reset pointer to small
continue;
}
else if(small[++j] == 0) // reached end of small
{
big[k++] = ' '; // replace chars from small with a single space
j = 0; // reset pointer to small
}
}
big[k] = '\0';
puts(big);
return 0;
}
or even better (no need for fixed sizes of strings):
#include <stdio.h>
int main() { // void main is deprecated...
char small[] = {"ol"};
char big[] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int k = 0; // pointer to written back big
for (i = 0; i < sizeof(big); i++)
{
if(small[j]!=big[i])
...
In C strings are terminated with a null character '\0'. Your code defines a somehow random number at the beginning (B and S) and iterates over that much characters instead of the exact number of characters, the strings actually contain. You can use the fact that the string is terminated by testing the content of the string in a while loop.
i = 0;
while (str[i]) {
...
i = i + 1;
}
If you prefer for loops you can write it also as a for loop.
for (i = 0; str[i]; i++) {
...
}
Your code does not move the contents of the remaining string to the left. If you replace two characters ol with one character , you have to move the remaining characters to the left by one character. Otherwise you would have a hole in the string.
#include <stdio.h>
int main() {
char small[] = "ol";
char big[] = "my older gradmom see my older sister";
int s; // index, which loops through the small string
int b; // index, which loops through the big string
int m; // index, which loops through the characters to be modified
// The following loops through the big string up to the terminating
// null character in the big string.
b = 0;
while (big[b]) {
// The following loops through the small string up to the
// terminating null character, if the character in the small
// string matches the corresponding character in the big string.
s = 0;
while (small[s] && big[b+s] == small[s]) {
// In case of a match, continue with the next character in the
// small string.
s = s + 1;
}
// If we are at the end of the small string, we found in the
// big string.
if (small[s] == '\0') {
// Now we have to modify the big string. The modification
// starts at the current position in the big string.
m = b;
// First we have to put the space at the current position in the
// big string.
big[m] = ' ';
// And next the rest of the big string has to be moved left. The
// rest of the big string starts, where the match has ended.
while (big[b+s]) {
m = m + 1;
big[m] = big[b+s];
s = s + 1;
}
// Finally the big string has to be terminated by a null
// character.
big[m+1] = '\0';
}
// Continue at next character in big string.
b = b + 1;
}
puts(big);
return 0;
}

Rearranging string letters

I was doing a program to copy all string words other than its first 2 words and putting a x at the end of it.
However i cant put x at its end. Please help!!!!
Below is my code.
#include<stdio.h>
#include<string.h>
int main()
{
char a[25], b[25];
int i, j, count = 0, l, k;
scanf("%[^\n]s", a);
i = strlen(a);
if (i > 20)
printf("Given Sentence is too long.");
else
{/* checking for first 2 words and counting 2 spaces*/
for (j = 0; j < i; j++)
{
if (a[j] == ' ')
count = count + 1;
if (count == 2)
{
k = j;
break;
}
}
/* copying remaining string into new one*/
for (j = 0; j < i - k; j++)
{
b[j] = a[j + k];
}
b[j + 1] = 'x';
printf("%s", b);
}
}
you are removing first two index. But you wrote k=j and if you check the current value j there it's 1. so you are updating wrongly k because you removed 2 indexes. So k value should be 2. So checked the below code
/* copying remaining string into new one*/
for (j = 0; j < i - 2; j++)
{
b[j] = a[j + 2];
}
b[j + 1] = 'x';
printf("%s", b);
Your index is off by one. After your second loop, the condition j < i-k was false, so j now is i-k. Therefore, the character after the end of what you copied is b[j], not b[j+1]. The correct line would therefore be b[j] = 'x';.
Just changing this would leave you with something that is not a string. A string is defined as a sequence of char, ending with a '\0' char. So you have to add b[j+1] = 0; as well.
After these changes, your code does what you intended, but still has undefined behavior.
One problem is that your scanf() will happily overflow your buffer -- use a field width here: scanf("%24[^\n]", a);. And by the way, the s at the and doesn't make any sense, you use either the s conversion or the [] conversion.
A somewhat sensible implementation would use functions suited for the job, like e.g. this:
#include<stdio.h>
#include<string.h>
int main(void)
{
// memory is *cheap* nowadays, these buffers are still somewhat tiny:
char a[256];
char b[256];
// read a line
if (!fgets(a, 256, stdin)) return 1;
// and strip the newline character if present
a[strcspn(a, "\n")] = 0;
// find first space
char *space = strchr(a, ' ');
// find second space
if (space) space = strchr(space+1, ' ');
if (space)
{
// have two spaces, copy the rest
strcpy(b, space+1);
// and append 'x':
strcat(b, "x");
}
else
{
// empty string:
b[0] = 0;
}
printf("%s",b);
return 0;
}
For functions you don't know, google for man <function>.
In C strings are array of chars as you know and the way C knows it is end of the string is '\0' character. In your example you are missing at the last few lines
/* copying remaining string into new one*/
for(j=0;j<i-k;j++)
{
b[j]=a[j+k];
}
b[j+1]='x';
printf("%s",b);
after the loop ends j is already increased 1 before it quits the loop.
So if your string before x is "test", it is like
't', 'e', 's', 't','\0' in char array, and since your j is increased more than it should have, it gets to the point just right of '\0', but characters after '\0' doesnt matter, because it is the end, so your x will not be added. Simple change to
b[j]='x';

int being read as pointer [duplicate]

This question already has answers here:
Variable Sized Arrays in C
(5 answers)
Closed 7 years ago.
I have to take a phrase from the user, and print it out in an inverted triangle with spaces in between. My program takes the phrase, stores it into an inputBuffer (array of char), and then creates a new array double the size of string. It fills up the first half with spaces, and the second half with the string itself. I want to print out the strLength char from the new array strLength times, just by shifting the range of strLength to (strLength*2-1) by 1 each time to the left. This ensures that in the first iteation, only the whole string is printed, second time one space at the beginning and one char at the end is not printed, as so on. Currently I am getting an error that even through strLength is an int variable, when I use it to create the new array it's apparently NOT a constant value.
int main(void) {
char inputBuffer[256];
char *pointer = inputBuffer;
char *temp = pointer;
int strLength = 0;
printf("enter your word: ");
scanf("%s", pointer);
//Calculate string length
while (*temp++) strLength++;
//Create an array double the size, first half for white spaces, and second half for the phrase.
char inputString[strLength * 2];
// ERROR: above expression inside the index must be a constant value.
int i, j;
//First half of array is number of spaces == number of char in phrase
for (i = 0; i < strLength; i++) {
inputString[i] = ' ';
}
//Reinitialize temp to use instead of pointer & put the string in the second half of inputString[]
temp = pointer;
for (j = 0; j < strLength; j++) {
inputString[i++] = *temp++;
}
//Just print the strLength indexes of inputStrng[] starting from half to end, and keep shifting the range by 1 position to the left.
for (i = strLength; i < (strLength * 2); i--) {
for (j = 0; j < strLength; j++) {
putchar(inputString[i + j]);
putchar(' ');
}
putchar('\n');
}
return 0;
}
The index variable your using to create inputString is infact not a constant, it's a variable..
If you want to create an array of a variable size, you have to use malloc.. You have to replace the line char inputString[strLength * 2]; with a malloc statement..
See this answer
So.. something like this:
char * inputString = malloc( sizeof(char) * ( (strLength * 2) + 1 ) );

code accounting for multiple delimiters isn't working

I have a program I wrote to take a string of words and, based on the delimiter that appears, separate each word and add it to an array.
I've adjusted it to account for either a ' ' , '.' or '.'. Now the goal is to adjust for multiple delimiters appearing together (as in "the dog,,,was walking") and still only add the word. While my program works, and it doesn't print out extra delimiters, every time it encounters additional delimiters, it includes a space in the output instead of ignoring them.
int main(int argc, const char * argv[]) {
char *givenString = "USA,Canada,Mexico,Bermuda,Grenada,Belize";
int stringCharCount;
//get length of string to allocate enough memory for array
for (int i = 0; i < 1000; i++) {
if (givenString[i] == '\0') {
break;
}
else {
stringCharCount++;
}
}
// counting # of commas in the original string
int commaCount = 1;
for (int i = 0; i < stringCharCount; i++) {
if (givenString[i] == ',' || givenString[i] == '.' || givenString[i] == ' ') {
commaCount++;
}
}
//declare blank Array that is the length of commas (which is the number of elements in the original string)
//char *finalArray[commaCount];
int z = 0;
char *finalArray[commaCount] ;
char *wordFiller = malloc(stringCharCount);
int j = 0;
char current = ' ';
for (int i = 0; i <= stringCharCount; i++) {
if (((givenString[i] == ',' || givenString[i] == '\0' || givenString[i] == ',' || givenString[i] == ' ') && (current != (' ' | '.' | ',')))) {
finalArray[z] = wordFiller;
wordFiller = malloc(stringCharCount);
j=0;
z++;
current = givenString[i];
}
else {
wordFiller[j++] = givenString[i];
}
}
for (int i = 0; i < commaCount; i++) {
printf("%s\n", finalArray[i]);
}
return 0;
}
This program took me hours and hours to get together (with help from more experienced developers) and I can't help but get frustrated. I'm using the debugger to my best ability but definitely need more experience with it.
/////////
I went back to pad and paper and kind of rewrote my code. Now I'm trying to store delimiters in an array and compare the elements of that array to the current string value. If they are equal, then we have come across a new word and we add it to the final string array. I'm struggling to figure out the placement and content of the "for" loop that I would use for this.
char * original = "USA,Canada,Mexico,Bermuda,Grenada,Belize";
//creating two intialized variables to count the number of characters and elements to add to the array (so we can allocate enough mmemory)
int stringCharCount = 0;
//by setting elementCount to 1, we can account for the last word that comes after the last comma
int elementCount = 1;
//calculate value of stringCharCount and elementCount to allocate enough memory for temporary word storage and for final array
for (int i = 0; i < 1000; i++) {
if (original[i] == '\0') {
break;
}
else {
stringCharCount++;
if (original[i] == ',') {
elementCount++;
}
}
}
//account for the final element
elementCount = elementCount;
char *tempWord = malloc(stringCharCount);
char *finalArray[elementCount];
int a = 0;
int b = 0;
//int c = 0;
//char *delimiters[4] = {".", ",", " ", "\0"};
for (int i = 0; i <= stringCharCount; i++) {
if (original[i] == ',' || original[i] == '\0') {
finalArray[a] = tempWord;
tempWord = malloc(stringCharCount);
tempWord[b] = '\0';
b = 0;
a++;
}
else {
tempWord[b++] = original[i];
}
}
for (int i = 0; i < elementCount; i++) {
printf("%s\n", finalArray[i]);
}
return 0;
}
Many issues. Suggest dividing code into small pieces and debug those first.
--
Un-initialize data.
// int stringCharCount;
int stringCharCount = 0;
...
stringCharCount++;
Or
int stringCharCount = strlen(givenString);
Other problems too: finalArray[] is never assigned a terminarting null character yet printf("%s\n", finalArray[i]); used.
Unclear use of char *
char *wordFiller = malloc(stringCharCount);
wordFiller = malloc(stringCharCount);
There are more bugs than lines in your code.
I'd suggest you start with something much simpler.
Work through a basic programming book with excercises.
Edit
Or, if this is about learning to program, try another, simpler programming language:
In C# your task looks rather simple:
string givenString = "USA,Canada Mexico,Bermuda.Grenada,Belize";
string [] words = string.Split(new char[] {' ', ',', '.'});
foreach(word in words)
Console.WriteLine(word);
As you see, there are much issues to worry about:
No memory management (alloc/free) this is handeled by the Garbage Collector
no pointers, so nothing can go wrong with them
powerful builtin string capabilities like Split()
foreach makes loops much simpler

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