strend function in C using pointers? - c

I have created a function for strend, which basically returns 1 if string t is present at the end of string s, however it never returns 1:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int strend(char *s, char *t) {
int p;
for (p = 0; p < strlen(s) - strlen(t); p++) {
*s++;
}
printf("%s\n%s\n", s, t);
if (s == t)
return 1;
return 0;
}
int main(void) {
int bool = strend("Hello", "ello");
printf("%i\n", bool);
return 0;
}
This gives me an output of:
ello
ello
0
So technically I should get 1. I assume the comparison using pointers is not used in this way?

You need to review your basic knowledge of C strings. There are lots of standard string functions in string.h that can help you with this test.
The basic problem is that the test s == t is valid, but you are comparing memory addresses here. You can see that is valid if you change the strings to test to
char test[] = "Hello";
int bool = strend_(test, test+1);
where test obviously is the same as your "Hello", and similarly, test+1 is the same as "ello" (try it by printing them). This correctly returns 1 with your routine.
In addition, I get two warnings:
on *s++; "warning: expression result unused [-Wunused-value]": you increment s but also ask what character is at that position through *s; and you don't use that information.
Fix by removing the * there.
on p < strlen(s) ..; "warning: comparison of integers of different signs: 'int' and 'unsigned long'", because strlen does not return a signed integer but an unsigned one (apparently, my header uses unsigned long).
Fix by declaring p as unsigned long, or even better, size_t.
Your entire routine can be condensed to a simple
int strend (char *s, char *t)
{
if (strlen(s) >= strlen(t) && !strcmp (s+strlen(s)-strlen(t),t))
return 1;
return 0;
}
It's not worth the trouble to cache the result of those four strlen calls into 2 temporary variables; a good compiler will work it out and do that for you. (A quick glance to the assembly output of the compiler I'm using – clang – shows it does, even with the default optimization settings.)
A slightly modified test, based on #M.M.'s comment:
int strend (char *s, char *t)
{
if (strlen(s) < strlen(t)) return 0;
return !strcmp (s+strlen(s)-strlen(t),t);
}
but attempting to optimize it this way is not as easy parsed as the routine above, and its assembly is ever so slightly "wordy" as well. Personally, I'd go for the more humanly readable version.

Use strcmp(3)
if (strcmp(s, t) == 0) return 1;
This actually compares the contents of the memory pointed to by s and t rather than their addresses.

Your code is broken in multiple ways:
The initial loop is a very cumbersome way to advance p by the difference of lengths if positive.
Once you have pointers at the same distance from the end of both strings, You should compare the characters with strcmp() (or memcmp() if you can first exclude the case of strlen(s) < strlen(t).
Comparing the pointers obtained after the loop will only work if t points inside the string pointed to by s, a special case that may or may not be produced by the compiler for the specific call in main: strend("Hello", "ello");.
Here is a modified version:
#include <string.h>
int strend(const char *str1, const char *str2) {
size_t len1 = strlen(str1);
size_t len2 = strlen(str2);
return len1 >= len2 && !memcmp(str1 + len1 - len2, str2, len2);
}

I corrected/modified your code, here is the code,
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
//#pragma warning(disable:4996)
int strend(char *s, char *t)
{
int p,flag=0,count=0;//count will be the starting index for *t
p = strlen(s) - strlen(t);//this will be the starting index for *s
while(count<strlen(t))
{
if (*(s+p) == *(t+count))
{
flag = 1;
count++;
p++;
continue;
}
else
{
flag = 0;
break;
}
}
return flag;
}
int main(void)
{
int flag = strend("Hello", "ello");
printf("%i\n", flag);
return 0;
}

This code works too.
#include <stdio.h>
#include <string.h>
int strend (char *s1, char *s2);
void main ()
{
char str1[20] = "somethings";
char str2[20] = "things";
int f;
f = strend (str1,str2);
if (f==1)
printf ("1");
else
printf ("0");
}
int strend (char *str1, char *str2)
{
int l = strlen(str1) - strlen(str2);
str1 = str1 + l;
int d = strcmp(str1,str2);
if (d == 0)
return 1;
else
return 0;
}

this code works well.
int strend(char *s, char *t){
while(*t & *s){
if(*t == *s){
t++;
}
s++;
}
return *t==*s;
}

Related

Converting a comma separated string to array

I have been trying to convert a string in array of integers, floats and characters. While I could get it work for integers and floats, there is some problem for characters.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *s1;
int k, no=5;
char* variable = "R1,R2,R3,R4,R5";
void* value;
s1 = calloc(no,sizeof(char)*81);
for (k=0; k<no; k++) s1[k] = strdup(mchar);
ListChar(variable, s1, no, ",");
memcpy(value, s1, no*sizeof(char)*81);
free(s1);
int i;
for (i = 0; i < no; i++)
printf("%s", value[i]);
printf("\n");
return 0;
}
In the header file I have
#define mchar "A...(81times)"
Implementation:
int ListChar(char *buf, char *list, int maxloop, char* delim)
{
int n = 0;
char *s,*t;
s= strdup(buf);
t= strtok(s,delim);
while ( t && (n<maxloop))
{
if (list!=NULL) list[n] =strdup(t);
n++;
t=strtok(NULL,delim);
}
free(s);
return(n);
}
During the calloc memory assignment when I watch s1 its 0xsomeadress ""
After the for loop s1 becomes 0xsomeadress "Garbage value 81 times"
When s1 is assigned to list its still reads the same garbage value.
And when list [n] = strdup(t) list[0] reads the first block of garbage value like -21 '\221 ṗ'.
t is getting delimited correctly. I even tried initializing char *s1[81] = {"something"} and looping it on j but it wont work, same problem, and I need to free s1 at the end because this function runs for number of times. I did it for integers and floats by list[n]=atoi(t) it works fine. Can anyone suggest me something?
There seems to be a fundamental misunderstanding about how strings work. Your s1 clearly needs to be a char ** and the usage of strdup is incorrect. If s1 is of type char *, then s1[k] is of type char. But strdup returns a char *, so s1[k] = strdup ... is clearly an error which your compiler ought to warn you about. Perhaps you want something like:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void * xmalloc(size_t s);
void
ListChar(const char *buf, char **list, int maxloop, int delim)
{
char set[] = {delim, 0};
for( int n = 0; n < maxloop; n += 1 ){
size_t len = strcspn(buf, set);
list[n] = xmalloc(len + 1);
memcpy(list[n], buf, len);
buf += len + 1;
}
}
int
main(int argc, char **argv)
{
int delim = ',';
(void)argc; /* Suppress compiler warning */
while( *++argv ){
char **s1;
int k, num = 1;
char *input = *argv;
for( const char *p = input; *p; p += 1 ){
if( *p == delim ){
num += 1;
}
}
s1 = xmalloc(num * sizeof *s1);
ListChar(input, s1, num, delim);
for( int i = 0; i < num; i += 1 ){
printf("%s\n", s1[i]);
}
free(s1);
}
return 0;
}
void *
xmalloc(size_t s)
{
void *rv = malloc(s);
if( rv == NULL ){
perror("malloc");
exit(EXIT_FAILURE);
}
return rv;
}
Note that the above code scans each string twice, which is not ideal. Rather than scanning the string to find the number of delimiters and then parsing the string, it would be better to do both in one pass. But for the purposes of demonstrating how to break up the string, that seems like unnecessary complexity. (Though it's actually simpler, IMO)

How to concatenate 2 strings using malloc and not the library functions

I need to create a function to concatenate 2 strings, in my case they are already given. I will need to concatenate the strings 'hello' and 'world!' to make it into 'helloworld!'. However, I can't use library functions besides strlen(). I also need to use malloc. I understand malloc would create n amounts of bytes for memory, however, how would I make it so that it can return a string array if thats possible.
Here is what I have so far,
#include <stdio.h>
#include <string.h>
int *my_strcat(const char* const str1, const char *const str2)
{
int s1, s2, s3, i = 0;
char *a;
s1 = strlen(str1);
s2 = strlen(str2);
s3 = s1 + s2 + 1;
a = char *malloc(size_t s3);
for(i = 0; i < s1; i++)
a[i] = str1[i];
for(i = 0; i < s2; i++)
a[i+s1] = str2[i];
a[i]='\0';
return a;
}
int main(void)
{
printf("%s\n",my_strcat("Hello","world!"));
return 0;
}
Thanks to anyone who can help me out.
This problem is imo a bit simpler with pointers:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *mystrcat(char *a, char *b) {
char *p, *q, *rtn;
rtn = q = malloc(strlen(a) + strlen(b) + 1);
for (p = a; (*q = *p) != '\0'; ++p, ++q) {}
for (p = b; (*q = *p) != '\0'; ++p, ++q) {}
return rtn;
}
int main(void) {
char *rtn = mystrcat("Hello ", "world!");
printf("Returned: %s\n", rtn);
free(rtn);
return 0;
}
But you can do the same thing with indices:
char *mystrcat(char *a, char *b) {
char *rtn = malloc(strlen(a) + strlen(b) + 1);
int p, q = 0;
for (p = 0; (rtn[q] = a[p]) != '\0'; ++p, ++q) {}
for (p = 0; (rtn[q] = b[p]) != '\0'; ++p, ++q) {}
return rtn;
}
Here is an alternate fix. First, you forgot #include <stdlib.h> for malloc(). You return a pointer to char from the function my_strcat(), so you need to change the function prototype to reflect this. I also changed the const declarations so that the pointers are not const, only the values that they point to:
char * my_strcat(const char *str1, const char *str2);
Your call to malloc() is incorrectly cast, and there is no reason to do so anyway in C. It also looks like you were trying to cast the argument in malloc() to size_t. You can do so, but you have to surround the type identifier with parentheses:
a = malloc((size_t) s3);
Instead, I have changed the type declaration for s1, s2, s3, i to size_t since all of these variables are used in the context of string lengths and array indices.
The loops were the most significant change, and the reason that I changed the consts in the function prototype. Your loops looked fine, but you can also use pointers for this. You step through the strings by incrementing a pointer, incrementing a counter i, and store the value stored there in the ith location of a. At the end, the index i has been incremented to indicate the location one past the last character, and you store a '\0' there. Note that in your original code, the counter i was not incremented to indicate the location of the null terminator of the concatenated string, because you reset it when you looped through str2. #jpw shows one way of solving this problem.
I changed main() just a little. I declared a pointer to char to receive the return value from the function call. That way you can free() the allocated memory when you are through with it.
Here is the modified code:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char * my_strcat(const char *str1, const char *str2)
{
size_t s1, s2, s3, i = 0;
char *a;
s1 = strlen(str1);
s2 = strlen(str2);
s3 = s1+s2+1;
a = malloc(s3);
while(*str1 != '\0') {
a[i] = *str1;
str1++;
i++;
}
while(*str2 != '\0') {
a[i] = *str2;
str2++;
i++;
}
a[i] = '\0'; // Here i = s1 + s2
return a;
}
int main(void)
{
char *str = my_strcat("Hello", "world!");
printf("%s\n", str);
/* Always free allocated memory! */
free(str);
return 0;
}
There are a few issues:
In the return from malloc you don't need to do any cast (you had the syntax for the cast wrong anyway) (see this for more information).
You need to include the header stdlib.h for the malloc function.
And most importantly, a[i]='\0'; in this i is not what you need it to be; you want to add the null char at the end which should be a[s3]='\0'; (the length of s1+s2).
This version should be correct (unless I missed something):
#include <stdio.h>
#include <stdlib.h> //for malloc
#include <string.h>
char *my_strcat(const char* const str1, const char *const str2)
{
int s1,s2,s3,i=0;
char *a;
s1 = strlen(str1);
s2 = strlen(str2);
s3 = s1+s2+1;
a = malloc(s3);
for(i = 0; i < s1; i++) {
a[i] = str1[i];
}
for(i = 0; i < s2; i++) {
a[i+s1] = str2[i];
}
a[s3-1] = '\0'; // you need the size of s1 + s2 + 1 here, but - 1 as it is 0-indexed
return a;
}
int main(void)
{
printf("%s\n",my_strcat("Hello","world!"));
return 0;
}
Testing with Ideone renders this output: Helloworld!

Adding String with Recursive function in C

I need to write a probram in C, which adds a string to a string etc. (for example '5' strings - It needs to read "vbvbvbvbvb" 5 times.) But it doesn't work? Help please!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char s[80];
int len;
int counter = 0;
char* repeat(char* s, int n) {
if (n > 0) {
if (s[len] == n) {
counter++;
}
len--;
repeat(s, (n++));
}
return s;
}
int main(void) {
printf("%s", repeat("vb", 5));
fflush(stdout);
return EXIT_SUCCESS;
}
You're trying to write into the end of "vb" which is a string in the constant pool. Don't do that. Allocate a string that is strlen(s) * n + 1 long and write into that.
Your base case is probably wrong. The base case should probably be when n == 0 which is when the empty string (nothing appended except terminating NUL as below) is appropriate.
Your recursive step (n++) should probably be (n - 1) to count down to that base case. As written, the post-increment does a useless assign and recurses with the same value of n.
I don't know what counter and len are supposed to do, but they looks redundant to me. len is uninitialized, so s[len] has undefined behavior.
After writing the n copies, you need to add a terminating NUL ('\0') at the end so that printf and similar functions can identify the end.
You are using s both as a global and a local variable, the function is working on the local.
Try not to use global variables where not necessary. Also, recursion is not necessary for this.
#include <stdio.h>
void concatenate_string(char *dest, const char *src, int n) {
char *s;
while(n--) {
s = (char*)src;
while(*(s))
*(dest++)=*(s++);
}
*(dest++) = 0;
}
int main(void) {
char out[80];
concatenate_string(out, "ab", 5);
printf("%s", out);
return 0;
}

Use pointer math instead of array indexing

I'm trying to solve a problem found on my C programming book.
#include <stdio.h>
char *f (char s[], char t[], char out[]);
int main(void)
{
char s[] = "ISBN-978-8884981431";
char t[] = "ISBN-978-8863720181";
char out[10];
printf ("%s\n", f(s,t,out));
return 0;
}
char *f (char s[], char t[], char out[]) {
int i;
for (i=0; s[i] == t[i]; i++)
out[i] = s[i];
out[i] = '\0';
return &out[0];
}
As you can see from the code, this code uses as return value &out[0]: does this means that the complete array is used as return value?
char *f (char s[], char t[], char out[]);
int main(void)
{
char s[] = "ISBN-978-8884981431";
char t[] = "ISBN-978-8863720181";
char out[10];
printf ("%s\n", f(s,t,out));
return 0;
}
char *f (char s[], char t[], char out[]) {
for (; *(s+=1) == *(t+=1);)
*(out+=1) = *s;
*(out+1) = '\0';
return out;
}
This is my proposed solution, but while the proposed code returns "ISBN-978-88", mine only returns "8".
The array is smaller than the lenght of the string, how the proposed code can work without any kind of overflow?
Thanks for your responses.
Your code is too aggressive on side effects: the += 1 operation (which is more commonly denoted simply as ++) should be applied after the copy to the output has been made, not after the comparison.
In addition, you need to save the value of the out buffer before incrementing the pointer, so that you could return a pointer to the beginning of the copied string.
char *orig = out;
for ( ; *s == *t ; s++, t++)
*out++ = *s;
*out = '\0';
return orig;
Demo on ideone.
Your code is returning a pointer to the end of the out array. Not the start. You need to stash the initial value of out and return that.
As an aside, the fact that you can do assignments inside a comparison doesn't mean it's a good idea. That code is going to be very hard to maintain.
&out[0] is equivalent to out. Since arrays in C are passed by reference, in a sense, yes it does return the entire array.
Your solution only prints "8" because you're returning a pointer into the middle of the array. When it tries to print the string, it has no way of knowing that it's in the middle of the array/string, thus you only get a substring printed.

Why does this program give segmentation fault?

It's a beginners question: Why is this breaking/giving an error?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *strtrim_right(char *p)
{
char *end;
int len;
len = strlen( p);
while (*p && len)
{
end = p + len-1;
if(isalpha(*end))
*end =0;
else
break;
}
return(p);
}
int main ()
{
char *x="PM123BFD";
strtrim_right(x);
printf("%s", x);
return 0;
}
Change
char *x="PM123BFD";
to
char x[]="PM123BFD";
You cannot modify a string literal, so instead pass the function a char array which it can modify.
I don’t see why it should break – I would rather expect an infinite loop: the while condition will always be true and the loop will never be left.
Rework the loop condition, it’s borked. Then look at the variables you have: you never change the values of either p or len. This isn’t right. Furthermore, the code inside the loop is much more complicated than need be. Think about whether you really need three variables here.
Ok thanks to the 2 answers above here is what seems to be ok now:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *strtrim_right(char *p)
{
char *end;
int len;
len = strlen( p);
while (*p && len)
{
end = p + len-1;
if(isalpha(*end))
*end =0;
else
break;
len = strlen(p);
}
return(p);
}
int main ()
{
char x[]="PM123BFD";
strtrim_right(x);
printf("%s", x);
return 0;
}

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