Wrong output in my stringreverse program. what is fault in my code? - c

#include<stdio.h>
#include<conio.h>
#include<string.h>
char* strreverse(char*);
int main()
{
char *rev_string;
char *name="computer";
clrscr();
rev_string=strreverse(name);
printf("%s", rev_string);
getch();
return 0;
}
char* strreverse(char *name)
{
int length=strlen(name);
char *ptr;
char *rstr;
for(ptr=name+(length-1);ptr>=name;ptr--)
{
*rstr=*ptr;
printf("%c",rstr);
rstr++;
}
*(rstr)=NULL;
return rstr;
}
the above is my code. i tried to write a program for string reverse without using arrays. But i am not getting the output retupmoc. what is wrong in my code? how to insert null char in char*?

#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <stdlib.h>
char* strreverse(const char*);
int main(){
char *rev_string;
char *name="computer";
clrscr();
rev_string=strreverse(name);
printf("%s\n", rev_string);
free(rev_string);
getch();
return 0;
}
char* strreverse(const char *name){
int length=strlen(name);
const char *ptr;
char *ret, *rstr = malloc(length + 1);
if(ret=rstr){
for(ptr=name+length;ptr != name;){
*rstr++ = *--ptr;
}
*rstr = '\0';
}
return ret;
}

You did not allocate memory to hold your reversed string. Try
char *rstr = calloc(1, length+1);
Also it should be
printf("%c", *rstr); // dereference
*(rstr)= '\0'; // instead of NULL

Here you find sweet and short solution for string reverse:
#include<stdio.h>
#include<string.h>
int strreverse(char* , char*);
int main()
{
char rev_string[10] = {0};
char name[10]="computer";
strreverse(name, rev_string);
printf("%s\n", rev_string);
return 0;
}
int strreverse(char *name, char *rStr)
{
int i = 0;
int length = strlen(name);
while(i < length)
{
rStr[i] = name[length-i-1];
i++;
}
return 0;
}
Try to run and have fun.

Related

"Warning: assignment makes integer from pointer without a cast" using a char malloc

I'm making a code to remove the file name and type from a path. However, i'm receiving warnings concerning the line where i change the content from a character. How could i get rid of the warning?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *getPath(char *fullPath){
char *aux;
int a, b, c;
aux = malloc(50 * sizeof(char));
aux = fullPath;
a = strlen(aux);
for(b=0; b<a; b++){
if (aux[b] == '/'){
c = b;
}
}
for(c; c < a; c++){
///PROBLEM HERE
aux[c] = "";
}
///PROBLEM HERE
return aux;
}
int main(void) {
char C[50];
char *path, *filename;
scanf("%s", C);
path = getPath(C);
printf("%s", path);
}
aux[c] = ""; // here "" is a char *
aux is a char *, therefore aux[c] is a char (not a string "")
aux[c] = '\0';
As written in the comments, there still have other mistakes in the rest of the code: for example aux value is erased.
Tried fixing the entire code.
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
bool getPath(char *const strippedPath, const int strippedPath_buflen,const char *const fullPath){
int b=strlen(fullPath);
for(;;){
--b;
if(b<0)
return false;
if('/'==fullPath[b])
break;
}
if(strippedPath_buflen<b+1)
return false;
strncpy(strippedPath,fullPath,b);
strippedPath[b]='\0';
return true;
}
int main(void) {
for(;;){
char C[50]={};
printf("> ");
fflush(stdout);
scanf("%s",C);
if(0==strcmp("quit",C))
break;
char path[3+1]={'X','X','X','X'};
if(getPath(path,4,C))
printf("%s\n",path);
else
printf("err\n");
}
return 0;
}
> aaaa/b.txt
err
> aaa/b.txt
aaa
> a/c/b.txt
a/c
> aa/b.txt
aa
> a/b.txt
a
> a/
a
> /b.txt
> b.txt
err
> quit

function of arrays of strings in c

This function stops awsering, and I can't spot the prob, can someone try to find it please?
It is suposed to give me the designed name and number from an array of strings.
#include <stdio.h>
#include <string.h>
int sameName();
char **getNumber();
char **getNumber (char *n[], char e[],int N){
int a;
for(a=0;a<N;a++){
if (sameName(n[a],e))
{
return n[a];
}
}
return "Not found!";
}
int sameName(char n[], char e[]){
int a;
for(a=0;e[a]!='\0';a++){
if (n[a]!=e[a])
{
return 0;
}
}
return 1;
}
int main (){
char numbers [5] [100] ={{"Ash 031"},{"Bomberman 021"},{"Rango 120"},{"Gigo Senhas 017"},{"Marcoreano 135"}};
char name [100];
char a [100];
scanf("%s",&a);
strcpy(name,getNumber (numbers,a,5));
printf("%s\n",name);
return 0;
}
You need to spicify the size of your two-dimension array when you pass it to your getNumber function. See http://c-faq.com/aryptr/pass2dary.html for details.
And for scanf you only need to pass a since it is a string.
Taking these plus fixing your prototypes as mentioned in the comment above, that gives :
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int sameName(char n[], char e[]);
char *getNumber (char n[][100], char e[],int N);
char *getNumber (char n[][100], char e[],int N){
int a;
for(a=0;a<N;a++){
if (sameName(n[a],e))
{
return n[a];
}
}
return "Not found!";
}
int sameName(char n[], char e[]){
int a;
for(a=0;e[a]!='\0';a++){
if (n[a]!=e[a])
{
return 0;
}
}
return 1;
}
int main (){
char numbers [5] [100] ={{"Ash 031"},{"Bomberman 021"},{"Rango 120"},{"Gigo Senhas 017"},{"Marcoreano 135"}};
char name [100];
char a [100];
scanf("%s",a);
strcpy(name,getNumber (numbers,a,5));
printf("%s\n",name);
return 0;
}

Getting wrong answer when counting array in c

#include <stdio.h>
#include <stdlib.h>
int countArrayChars(char *strArray[]){
int i=0;
while (strArray[i] != '\0'){
i++;
}
printf("%d\n", i);
return i;
}
int main(int argc, const char * argv[]) {
char *dog[] = {"dog"};
countArrayChars(dog);
For some reason, it prints "5".
Shouldn't it print 3?
I even tried to put \0 after the "g".
You declare array of string and initialize it with dog.
char *dog[] = {"dog"};
Actually it represented as
dog[0] = "Dog"; //In your case only element index with 0.
...............
...............
dog[n] = "tiger"; //If there Have n+1 element
Hence your array size is 1. Which hold constant string dog. To access it you should use dog[0].
So without less modification you can use your code as:
int countArrayChars(char *strArray[])
{
int i=0;
while (strArray[0][i] != '\0')
{
i++;
}
printf("%d\n", i);
return i;
}
int main(int argc, const char * argv[])
{
char *dog[] = {"dog"};
countArrayChars(dog);
}
Or if you want to declare a string use
char *dog = "dog";
or
char dog[] = "dog";
Please try this
#include <stdio.h>
#include <stdlib.h>
int countArrayChars(char *strArray){
int i=0;
while (strArray[i] != '\0'){
i++;
}
printf("%d\n", i);
return i;
}
int main(int argc, const char * argv[]) {
char *dog[] = "dog";
countArrayChars(dog);
}

Cut out section of a string with another string

I got this part of a C program:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void){
char *mRNA = spleissen("AUAGUAAAAGCUCUGUUUAGGAGA", "GU", "AG");
printf("mRNA: %s\n", mRNA);
free(mRNA);
return 0;
}
I have to write the function spleissen which should work like this: it cuts out a string which goes from a GU to an AG and everything in between those two. So the program output is:
mRNA: AUACUCUGAGA
I don't really know how I can cut those parts out.
I am not allowed to use includes other than stdio, string and stdlib.
char *spleissen(const char *src, const char *start, const char *end){
size_t len = strlen(src);
char *s, *e, *ret, *work;
ret = work = malloc(len + 1);
strcpy(work, src);
len = strlen(end);
while(s = strstr(work, start)){
if((e = strstr(s, end))==NULL)
break;//delete upto last?
memmove(s, e + len, strlen(e+len)+1);
work = s;
}
return ret;
}
I think you can simply do this:
char *spleissen(char *array, char *G, char *A)
{
int l=strlen(array);
int i, j=0;
char returnstr[10010];
int b=0;
for(i=0; i<l; i++)
{
if(G[0]==array[i] && G[1]==array[i+1])
{
b=1, i++;
continue;
}
else if(A[0]==array[i] && A[1]==array[i+1] && b==1)
{
b=0, i++;
continue;
}
if(b==0)
{
returnstr[j]=array[i];
j++;
}
}
return returnstr;
}

how to return a char array from a function in C

I want to return a character array from a function. Then I want to print it in main. how can I get the character array back in main function?
#include<stdio.h>
#include<string.h>
int main()
{
int i=0,j=2;
char s[]="String";
char *test;
test=substring(i,j,*s);
printf("%s",test);
return 0;
}
char *substring(int i,int j,char *ch)
{
int m,n,k=0;
char *ch1;
ch1=(char*)malloc((j-i+1)*1);
n=j-i+1;
while(k<n)
{
ch1[k]=ch[i];
i++;k++;
}
return (char *)ch1;
}
Please tell me what am I doing wrong?
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char *substring(int i,int j,char *ch)
{
int n,k=0;
char *ch1;
ch1=(char*)malloc((j-i+1)*1);
n=j-i+1;
while(k<n)
{
ch1[k]=ch[i];
i++;k++;
}
return (char *)ch1;
}
int main()
{
int i=0,j=2;
char s[]="String";
char *test;
test=substring(i,j,s);
printf("%s",test);
free(test); //free the test
return 0;
}
This will compile fine without any warning
#include stdlib.h
pass test=substring(i,j,s);
remove m as it is unused
either declare char substring(int i,int j,char *ch) or define it before main
Lazy notes in comments.
#include <stdio.h>
// for malloc
#include <stdlib.h>
// you need the prototype
char *substring(int i,int j,char *ch);
int main(void /* std compliance */)
{
int i=0,j=2;
char s[]="String";
char *test;
// s points to the first char, S
// *s "is" the first char, S
test=substring(i,j,s); // so s only is ok
// if test == NULL, failed, give up
printf("%s",test);
free(test); // you should free it
return 0;
}
char *substring(int i,int j,char *ch)
{
int k=0;
// avoid calc same things several time
int n = j-i+1;
char *ch1;
// you can omit casting - and sizeof(char) := 1
ch1=malloc(n*sizeof(char));
// if (!ch1) error...; return NULL;
// any kind of check missing:
// are i, j ok?
// is n > 0... ch[i] is "inside" the string?...
while(k<n)
{
ch1[k]=ch[i];
i++;k++;
}
return ch1;
}
Daniel is right: http://ideone.com/kgbo1C#view_edit_box
Change
test=substring(i,j,*s);
to
test=substring(i,j,s);
Also, you need to forward declare substring:
char *substring(int i,int j,char *ch);
int main // ...

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