I am trying to write a program that counts the amount a single word occurs in the program input. I was asked to use the strncmp() and strnlen() functions
This is my code.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) {
char *p, *s = "let", *e = "#EOF";
p = malloc(sizeof(char) * 1024);
int i = 0;
scanf("%1023s", p);
for (; strncmp(p, e, 4) != 0; ++p) {
if (strncmp(p, s, 3) == 0)
i++;
}
printf("\n%d", i);
}
The idea was to store the input string (which is given not to be larger then 1023 Char) at the pointer location p. And to then iterate from p until I find the string "#EOF", which should be given at the end of each input for my program. This is a part of my assignment. Through each iteration I am checking if the word "let" is in the next 3 characters of my string. Overlapping words should be counted aswel.
This is my program output when I input:
Let it be, let it be, let it be, let it be.
Whisper words of wisdom, let it be.
#EOF
Let it be, let it be, let it be, let it be.
PS C:\Users\*****\Downloads\C code> Whisper words of wisdom, let it be.
Whisper : The term 'Whisper' is not recognized as the name of a cmdlet, function, script file, or operable program. Check the spelling of the name, or if a path was included,
verify that the path is correct and try again.
At line:1 char:1
+ Whisper words of wisdom, let it be.
+ ~~~~~~~
+ CategoryInfo : ObjectNotFound: (Whisper:String) [], CommandNotFoundException
+ FullyQualifiedErrorId : CommandNotFoundException
Which indicates that my program terminates before it has read the entire string. A friend of mine had this code
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *z;
char *s = "let";
char *str;
int i = 0;
int main(void) {
str = malloc(sizeof(char) * 1024);
scanf("%1023[^#EOF]s", str);
for (z = str; *z; z++) {
printf("%s", z);
if (strncmp(z, s, 3) == 0)
i++;
}
printf("%d", i);
return 0;
}
Where the condition of the for loop is what p points to. It terminates when p points to '/0' which is at the end of the string. Is this a good explanation?
Related
I am trying to print each word in a single line of a given sentence. It worked perfectly fine but a '_' appears in end of line. please help me with it and also proper manar to write it.
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
char *s,i,check=0;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s);
s = realloc(s, strlen(s) + 1);
for(i=0;i<1024;i++ ||check<=2)
{
if(*(s+i)!=' ')
{
printf("%c",*(s+i));
check=0;
}
else
{
printf("\n");
check++;
}
// fflush(stdin);
}
return 0;
}
Output:
dkf fja fjlak d
dkf
fja
fjlak
d SER_
Output2:
-for(i=0;i<20;i++ ||check<=2)-
hello I am suraj Ghimire
hello
I
am
suraj
Ghi
I am not sure your code works as you say..
The type of i is not a char *, so it should be int.
You process the input string without considering the NULL terminating char, which leads to a lot of garbage prints.
You do not release allocated memory.
I suggest this slightly modified version:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
char *s, *p;
/* Allocate a new string and verify the allocation has succeeded. */
s = malloc(1024 * sizeof(char));
if (!s) {
printf("malloc failed\n");
return 1;
}
/* Read from user. */
scanf("%[^\n]", s);
/* Work on a copy of `s` (must simpler and faster than a indexed access). */
p = s;
while (*p) {
if (*p != ' ') {
printf("%c",*p);
}else{
printf("\n");
}
p++;
}
free(s);
return 0;
}
Example output:
$ ./a.out
abc def gh i j kmlm opqrst
abc
def
gh
i
j
kmlm
opqrst
EDIT: As requested by the OP, further details regarding the NULL terminating char.
By convention, strings (array of characters) end with a specific character which we call the NULL terminating char. This character is 0 and marks the end of the string data.
In your example, the buffer which store the string is dynamically allocated in RAM. If you do not check for the NULL terminating character of the string, then you keep processing data as if it is part of the string (but it is not).
Going beyond this character make you access the following memory data (which is part of your program RAM data). Since these data can be anything (ranging from 0 to 255), printing them may lead to "gibberish" because they may not be printable and are definitely not consistent with your string.
In the "best" case the program halts with a "segmentation fault" because you are accessing a memory region you are not allowed to. In the "worst" case you print a lot of things before crashing.
This is typically called a data leak (whether it is RAM or ROM) because it exposes internal data of your program. In the specific case of your example there no sensitive data. But! Imagine you leak passwords or private keys stored in your program .. this can be a severe security issue!
There are a couple issues with your code.
Firstly, you need to check that the for loop does not exceed the bounds of the string.
Your for loop is always set to true because the logical OR operator || has a higher precedence than the comma operator. Because of this the loop will always run unless it gets stopped with break
Lastly your check is never reset to 0 after it reaches a value of 2.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main() {
char *s,i,check=0;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s);
s = realloc(s, strlen(s) + 1);
for(i=0; i<strlen(s); i++) {
if(*(s+i) != ' ') {
printf("%c",*(s+i));
check=0;
} else {
printf("\n");
check++;
if (check > 2) break;
}
}
return 0;
}
Output:
Hello, this is a test
Hello,
this
is
a
test
for(i=0;i<1024;i++ ||check<=2)
There are two issues. One is length of string won't always be 1024, so it might be good to determine the length of string before print the string. The other is check<=2, which have to put in the second part of the for loop, so the test will be evaluated. Also it is better to calculate the length of string once. So I store the length of string in len.
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *s, i, check = 0;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s);
s = realloc(s, strlen(s) + 1);
size_t len = strlen(s);
for (i = 0; i < len || check <= 2; i++) {
if (*(s + i) != ' ') {
printf("%c", *(s + i));
check = 0;
} else {
printf("\n");
check++;
}
// fflush(stdin);
}
return 0;
}
I'm still a newbie to C so please forgive me if anything below is wrong. I've searched this up online but nothing really helped.
Right now, I have the following code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void appendStr (char *str, char c)
{
for (;*str;str++); // note the terminating semicolon here.
*str++ = c;
*str++ = 0;
}
int main(){
char string[] = "imtryingmybest";
char result[] = "";
for(int i = 0; i < strlen(string); i++){
if(i >= 0 && i <= 3){
appendStr(result, string[i]);
}
}
printf("%s", result);
}
Basically, I'm trying to add the first 4 characters of the String named string to result with a for loop. My code above did not work. I've already tried to use strcat and strncat and neither of them worked for me either. When I used
strcat(result, string[i]);
It returns an error saying that the memory cannot be read.
I know that in this example it might have been easier if I just did
appendStr(result, string[0]);
appendStr(result, string[1]);
appendStr(result, string[2]);
appendStr(result, string[3]);
But there is a reason behind why I'm using a for loop that couldn't be explained in this example.
All in all, I'd appreciate it if someone could explain to me how to append individual characters to a string in a for loop.
The following code doesnt use your methods but successfully appends the first 4 chars to result
#include <stdio.h>
#include <string.h>
int main()
{
// declare and initialize strings
char str[] = "imtryingmybest";
char result[5]; // the 5th slot is for \0 as all strings are null terminated
// append chars to result
strncat(result, str, 4);
// ^ ^ ^
// | | |- number of chars to be appended
// | | - string to be appended from
// | - string to append to
// print string
printf("result: %s\n", result);
return (0);
}
The result of the above is as wanted:
>> gcc -o test test.c
>> ./test
result: imtr
Let me know if anything is not clear so i can elaborate further
string was ruined by the overflow of result buffer.
appendStr can be executed only once. next time strlen(string) will return 0. because *str++ = 0; has been written to the space of string.
result buffer has only 1 byte space, but you write 2 byte to it in appendStr call.
the second byte will ruin string space.
I suggest debug with gdb.
try to get rid of Magic numbers
#define BUFF_SIZE 10 // define some bytes to allocate in result char array
#define COPY_COUNT 4 // count of chars that will be copied
int main(){
char string[] = "imtryingmybest";
char result[BUFF_SIZE] {}; // allocate some chunk of memory
for(size_t i = 0; i < strlen(string); i++){
if(i < COPY_COUNT){
appendStr(result, string[i]);
}
}
printf("%s", result);
}
I showed the solution code Paul Yang showed the problem
As others have pointed out the code has a simple mistake in the allocation of the destination string.
When declaring an array without specifying its size, the compiler deduces it by its initializer, which in your case means a 0 + the NULL character.
char result[] = ""; // means { '\0' };
However, I think that the bigger issue here is that you're effectively coding a Schlemiel.
C strings have the serious drawback that they don't store their length, making functions that have to reach the end of the string linear in time complexity O(n).
You already know this, as shown by your function appendStr()
This isn't a serious issue until start you appending characters or strings in a loop.
In each iteration of your loop appendStr() reaches the last character of the string, and extends the string, making the next iteration a little slower.
In fact its time complexity is O(n²)
Of course this is not noticeable for small strings or loops with few iterations, but it'll become a problem if the data scales.
To avoid this you have to take into account the growing size of the string.
I modified appendStr() to show that now it starts from the last element of result
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void appendStr (char *str, char c, char *orig)
{
printf("i: %ld\n", str - orig);
for (; *str; str++); // note the terminating semicolon here.
*str++ = c;
*str++ = 0;
}
int main()
{
char string[32] = "imtryingmybest";
char result[32] = "";
for(int i = 0; i < strlen(string); i++) {
if(i >= 0 && i <= 3) {
// I'm passing a pointer to the last element of the string
appendStr(&result[i], string[i], result);
}
}
printf("%s", result);
}
You can run it here https://onlinegdb.com/HkogMxbG_
More on Schlemiel the painter
https://www.joelonsoftware.com/2001/12/11/back-to-basics/
https://codepen.io/JoshuaVB/pen/JzRoyp
I am trying to run the following code, but during execution, the code does not go into the if condition.
Why does the code not enter the if condition during runtime? I have marked the problem condition.
Running this program on Windows 10.
Thread model: posix
gcc version 5.1.0 (tdm64-1)
I have tried using the ternary operator and the if statement with a different string, and strchr works fine in that case.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void main() {
static char str[] = "hello world";
static char inputTime[] = "12:05:10PM";
char *result = strchr(str, 'w');
long int tempNum = 0;
char *token, tempStr[10], delimit[] = ":";
if (strchr(str, 'w'))
printf("\nFound w");
else
printf("\nDid not find w");
(strchr(inputTime, 'P')) ? printf("\nTrue") : printf("\nFalse");
token = strtok(inputTime, delimit);
if (strchr(inputTime, 'P')) {
printf("Found PM\n");
tempNum = strtol(token, NULL, 10);
if (tempNum != 12)
tempNum += 12;
sprintf(tempStr, "%lu", tempNum);
}
printf("\ntempStr: %s", tempStr);
}
The above code gives me this output:
C:\Users\XX\Documents\Tests\c-programming>a.exe
Found w
True
tempStr: σ#
The strtok function splits the given input string into tokens. It does this by modifying the string to tokenize, placing a null byte in place of the delimiter to search for.
So after the call to strtok, inputTime looks like this:
{ '1','2','\0','0','5',':','1','0','P','M','\0' }
A null byte is put in place of the first :. So if you were to print inputTime you would get 12, meaning you won't find a P.
Because the input string is modified, you should search for P before calling strtok.
I tried using strncmp but it only works if I give it a specific number of bytes I want to extract.
char line[256] = This "is" an example. //I want to extract "is"
char line[256] = This is "also" an example. // I want to extract "also"
char line[256] = This is the final "example". // I want to extract "example"
char substring[256]
How would I extract all the elements in between the ""? and put it in the variable substring?
Note: I edited this answer after I realized that as written the code would cause a problem as strtok doesn't like to operate on const char* variables. This was more an artifact of how I wrote the example than a problem with the underlying principle - but apparently it deserved a double downvote. So I fixed it.
The following works (tested on Mac OS 10.7 using gcc):
#include <stdio.h>
#include <string.h>
int main(void) {
const char* lineConst = "This \"is\" an example"; // the "input string"
char line[256]; // where we will put a copy of the input
char *subString; // the "result"
strcpy(line, lineConst);
subString = strtok(line,"\""); // find the first double quote
subString=strtok(NULL,"\""); // find the second double quote
printf("the thing in between quotes is '%s'\n", subString);
}
Here is how it works: strtok looks for "delimiters" (second argument) - in this case, the first ". Internally, it knows "how far it got", and if you call it again with NULL as the first argument (instead of a char*), it will start again from there. Thus, on the second call it returns "exactly the string between the first and second double quote". Which is what you wanted.
Warning: strtok typically replaces delimiters with '\0' as it "eats" the input. You must therefore count on your input string getting modified by this approach. If that is not acceptable you have to make a local copy first. In essence I do that in the above when I copy the string constant to a variable. It would be cleaner to do this with a call to line=malloc(strlen(lineConst)+1); and a free(line); afterwards - but if you intend to wrap this inside a function you have to consider that the return value has to remain valid after the function returns... Because strtok returns a pointer to the right place inside the string, it doesn't make a copy of the token. Passing a pointer to the space where you want the result to end up, and creating that space inside the function (with the correct size), then copying the result into it, would be the right thing to do. All this is quite subtle. Let me know if this is not clear!
if you want to do it with no library support...
void extract_between_quotes(char* s, char* dest)
{
int in_quotes = 0;
*dest = 0;
while(*s != 0)
{
if(in_quotes)
{
if(*s == '"') return;
dest[0]=*s;
dest[1]=0;
dest++;
}
else if(*s == '"') in_quotes=1;
s++;
}
}
then call it
extract_between_quotes(line, substring);
#include <string.h>
...
substring[0] = '\0';
const char *start = strchr(line, '"') + 1;
strncat(substring, start, strcspn(start, "\""));
Bounds and error checking omitted. Avoid strtok because it has side effects.
Here is a long way to do this: Assuming string to be extracted will be in quotation marks
(Fixed for error check suggested by kieth in comments below)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(){
char input[100];
char extract[100];
int i=0,j=0,k=0,endFlag=0;
printf("Input string: ");
fgets(input,sizeof(input),stdin);
input[strlen(input)-1] = '\0';
for(i=0;i<strlen(input);i++){
if(input[i] == '"'){
j =i+1;
while(input[j]!='"'){
if(input[j] == '\0'){
endFlag++;
break;
}
extract[k] = input[j];
k++;
j++;
}
}
}
extract[k] = '\0';
if(endFlag==1){
printf("1.Your code only had one quotation mark.\n");
printf("2.So the code extracted everything after that quotation mark\n");
printf("3.To make sure buffer overflow doesn't happen in this case:\n");
printf("4.Modify the extract buffer size to be the same as input buffer size\n");
printf("\nextracted string: %s\n",extract);
}else{
printf("Extract = %s\n",extract);
}
return 0;
}
Output(1):
$ ./test
Input string: extract "this" from this string
Extract = this
Output(2):
$ ./test
Input string: Another example to extract "this gibberish" from this string
Extract = this gibberish
Output(3):(Error check suggested by Kieth)
$ ./test
Input string: are you "happy now Kieth ?
1.Your code only had one quotation mark.
2.So the code extracted everything after that quotation mark
3.To make sure buffer overflow doesn't happen in this case:
4.Modify the extract buffer size to be the same as input buffer size
extracted string: happy now Kieth ?
--------------------------------------------------------------------------------------------------------------------------------
Although not asked for it -- The following code extracts multiple words from input string as long as they are in quotation marks:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(){
char input[100];
char extract[50];
int i=0,j=0,k=0,endFlag=0;
printf("Input string: ");
fgets(input,sizeof(input),stdin);
input[strlen(input)-1] = '\0';
for(i=0;i<strlen(input);i++){
if(input[i] == '"'){
if(endFlag==0){
j =i+1;
while(input[j]!='"'){
extract[k] = input[j];
k++;
j++;
}
endFlag = 1;
}else{
endFlag =0;
}
//break;
}
}
extract[k] = '\0';
printf("Extract = %s\n",extract);
return 0;
}
Output:
$ ./test
Input string: extract "multiple" words "from" this "string"
Extract = multiplefromstring
Have you tried looking at the strchr function? You should be able to call that function twice to get pointers to the first and second instances of the " character and use a combination of memcpy and pointer arithmetic to get what you want.
How can you code this in C language if the output is like this? I need strings format of the code because our topic is strings.
#include <stdio.h>
#include <stdlib.h>
void main()
{
char my_string[50];
printf("Enter a word:");
scanf("%s", my_string);
printf("Enter a word:");
scanf("%s", my_string);
// Some unknown code here...
// this part is my only problem to solve this.
getch();
}
Output:
Hello -> (user input)
World -> (user input)
HWeolrllod -> (result)
Okay, you need to do some investigating. We don't, as a general rule, do people's homework for them since:
it's cheating.
you'll probably get caught out if you copy verbatim.
it won't help you in the long run at all.
The C library call for user input that you should use is fgets, along the line of:
char buffer[100];
fgets (buffer, sizeof(buffer), stdin);
This will input a string into the character array called buffer.
If you do that with two different buffers, you'll have the strings in memory.
Then you need to create pointers to them and walk through the two strings outputting alternating characters. Pointers are not an easy subject but the following pseudo-code may help:
set p1 to address of first character in string s1
set p1 to address of first character in string s1
while contents of p1 are not end of string marker:
output contents of p1
add 1 to p1 (move to next character)
if contents of p2 are not end of string marker:
output contents of p2
add 1 to p2 (move to next character)
while contents of p2 are not end of string marker:
output contents of p2
add 1 to p2 (move to next character)
Translating that into C will take some work but the algorithm is solid. You just need to be aware that a character pointer can be defined with char *p1;, getting the contents of it is done with *p1 and advancing it is p = p + 1; or p1++;.
Short of writing the code for you (which I'm not going to do), there's probably not much else you need.
void main()
{
char my_string1[50],my_string2[50]; int ptr;
ptr=0;
printf("Enter a word : ");
scanf("%s",my_string1);
printf("enter a word");
scanf("%s",my_string2);
while(my_string1[ptr]!='\0' && my_string2[ptr]!='\0')
{
printf("%c%c",my_string1[ptr],my_string2[ptr]);
ptr++;
}
if(my_string1[ptr]!='\0')
{
while(my_string1[ptr]!='\0')
{ printf("%c",my_string1[ptr]);
ptr++;
}
}
else
{
while(my_string2[ptr]!='\0')
{printf("%c",my_string2[ptr]);
ptr++;
}
}
}
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void main()
{
char my_string1[50],my_string2[50];
int i,l1=1,l2=0;
printf("Enter a word:");
scanf("%s", my_string1);
printf("Enter a word:");
scanf("%s", my_string2);
l1=strlen(my_string1); /* Length of 1st string */
l2=strlen(my_string2); /* Length of 2nd string */
if(l1==l2)
{
for(i=0;i<l1;i++)
{
printf("%c%c",my_string1[i],my_string2[i]);
}
}
else
{
printf("Length of the entered strings do not match");
}
}
This is your required code.
You can see that output needs to be a String containing all chars of User String1 and User String2 one by one...
You can do this like...
//add #include<String.h>
int l1=strlen(s1);
int l2=strlen(s2);
if(l1!=l2)
{
printf("length do not match");
return 0;
}
char ansstr[l1+l2];
int i,j=0,k=0;
for(i=0;i<l1+l2;i=i+2)
{
ansstr[i]=s1[j];
ansstr[i+1]=s2[k];
j++;
k++;``
}
//ansstr is your answer
Ok, here's your code. Come on guys, if he asked here it means he can't solve this.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
char str1[] = "abcdefghijklmopq";
char str2[] = "jklm";
int len1 = strlen(str1);
int len2 = strlen(str2);
int c1 = 0, c2 = 0;
int max = (len1 > len2) ? len1 : len2 ;
char *result = malloc(len1 + len2);
for(c1 = 0; c1 <= max; c1++) {
if(c1 < len1)
result[c2++] = str1[c1];
if(c1 < len2)
result[c2++] = str2[c1];
}
result[c2] = 0;
printf("\n%s\n", result);
return 0;
}
Basically the loop picks up a character from str1 and appends it to result. Then it picks a character, which stands in the same position as the first from str2 and appends it to result, just as before. I increment c2 by 2 every time because I'm adding 2 chars to result. I check if c1 is bigger that the length of the strings because I want to copy only the characters in the string without the terminating \0. If you know that your strings have the same length you can omit these ifs.