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;
}
Related
#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);
}
I want to pass an array of characters i.e. a String in c
int main()
{
const char c[]="Joseph";
TestWord(&c,&c);
return 0;
}
int TestWord(char tiles[], char word[])
{
return tiles;
}
#include <stdio.h>
char *TestWord(char tiles[], char word[]);
int main()
{
char c[]="Joseph";
char r;
r = *TestWord(c,c);
return 0;
}
char *TestWord(char tiles[], char word[])
{
return tiles;
}
You pass through the arrays without the & as arrays don't need those, as they are already somewhat like pointers, just like how you would scanf an array without the & symbol.
Don't forget that if you are returning tiles that you should save that in a variable.
you could pass a string(character array) in C in many ways.
This code passes the string a to the function PRINT. Note that in this method the base address of the array is sent to the function.
#include<stdio.h>
void PRINT(char b[])
{
printf("%s",b);
}
int main()
{
char a[]="hello";
PRINT(a);
return 0;
}
#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.
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 // ...
I'm a beginner in C language. After reading the initial chapters of Ritchie's book, I wrote a program to generate random numbers and alphabets.
The program compiles fine with gcc. However on running it, it gives an error "Segmentation fault", which is incomprehensible to my limited knowledge. I'd be glad to understand what I've written wrong.
#include <stdio.h>
#include <stdlib.h>
#include "conio.h"
#include <time.h>
long int genrandom(int,int);
void randAlph(void);
char letterize(int);
int main (void) {
// char full[9];
// char part_non[4];
srand(time(0));
int i;
for (i=0;i<50;++i) {
randAlph();
};
}
long int genrandom(int mino,int maxo) {
int val=mino+rand()/(RAND_MAX/(maxo-mino)+1);
return val;
}
void randAlph (){
int val;
char text;
val=genrandom(0,26);
// return val;
text=letterize(val);
printf("%s ,",text);
}
char letterize(int num) {
char letter='A'+num;
return letter;
}
printf("%s ,",text); is wrong - it says that text is a nul-terminated array of chars. Use
printf("%c ,", text);
instead to print your single char.
#include <stdio.h>
#include <stdlib.h>
#include "conio.h"
#include <time.h>
int genrandom(int,int);
void randAlph(void);
char letterize(int);
int main (void) {
// char full[9];
// char part_non[4];
srand(time(0));
int i;
for (i=0;i<50;++i) {
randAlph();
};
}
int genrandom(int mino,int maxo) {//changed function return type to int
int val=mino+rand()/(RAND_MAX/(maxo-mino)+1); //Be careful when you are using '/' operator with integers
return val; //returning int here why set return type to long int?
}
void randAlph (){
int val;
char text;
val=genrandom(0,26);
// return val;
text=letterize(val);
printf("%c ,",text);//Replace %s with %c
}
char letterize(int num) { //No bound checking on num eh?
char letter='A'+num;
return letter;
}
That's all I had to say. :)
Why use %s when text is char. You dont need a string type in the function. Just a char would do. Change in the function : void randAlph ()
printf("%s ,",text);
to
printf("%c ,", text);