int check(int i,int j,char test);
int main(int argc, char *argv[])
{
char mat[5][5];
char *anahtar;
anahtar=(char*)malloc (length*sizeof(char));
//i take length from user with scanf
int k=0;
if (check(i,j,anahtar[k])==1)
{
mat[i][j]=anahtar[k];
}
int check(int i,int j,char test)
{
int a=0;
int b=0;
if (mat[a][b]==test)
{
return 1;
}
else
{
return 0;
}
}
}
It gives error
undefined reference to `check'|
anahtar[] is a char array.So why cant i pass anahtar[k] in argument?
I already have protoype. PRoblem is not that.
The problem is that your check function is inside the main. Place it outside of main.
Also do not cast malloc;
anahtar = malloc (length*sizeof(char));
You need to declare it before you use it or
put this beforehand
int check(int i,int j,char test);
Related
I'm writing a program that can modify rows and cols from a function of a function. I don't understand how to do pointer of a pointer.
void changeNum(int*setRows, int *setCol)
{
changeNum2(*setRows,*setCol);
}
void changeNum2(int*setRows, int *setCol)
{
*setRows=5;
*setCol=5;
}
int main() {
int*row=10;
int* col=10;
changeNum(&row,&col);
printf("%d %d",row,col);
return 0;
}
First
int*row=10;
int* col=10;
This is wrong. Assigning hardcoded address. You don't want this
int row=10;
int col=10;
How to get the address of the row and col?
&row and &col.
How to pass it to function?
Call it, changeNum(&row,&col);
void changeNum(int*setRows, int *setCol)
{
...
}
How to pass pointer to pointer?
void changeNum(int*setRows, int *setCol)
{
chnageNum2(&setRows, &setCol);
}
ChangeNum2 how it would change value?
void chnageNum2(int **setRows, int **setCol){
**setRows = 110;
**setCol = 110;
}
Can we do the same change using changeNum() only?
Yes we can do that.
void changeNum(int*setRows, int *setCol)
{
*setRows = 110;
*setCol = 110;
}
Definitely check this. Grab a book. It will help a lot.
The complete code will be
void changeNum(int*setRows, int *setCol)
{
changeNum2(&setRows,&setCol);
}
void changeNum2(int**setRows, int **setCol)
{
**setRows=5;
**setCol=5;
}
int main(void) {
int row=10;
int col=10;
changeNum(&row,&col);
printf("%d %d",row,col);
return 0;
}
#include <stdio.h>
void changeNum(int*, int*);
void changeNum2(int*, int*);
void changeNum(int* setRows, int *setCol) {
changeNum2(setRows,setCol);
}
void changeNum2(int* setRows, int *setCol) {
*setRows=5;
*setCol=5;
}
int main() {
int row=10;
int col=10;
changeNum(&row, &col);
printf("%d %d\n", row, col);
return 0;
}
It takes sometime to grasp that every C function parameter is passed by value. So you can safely pass setRows pointer to the second function simply by its value.
Also, it's necessary to declare previously the function changeNum2, I've included the declaration without parameter names to clarify it's possible.
I strongly recommend reading this book: https://en.wikipedia.org/wiki/The_C_Programming_Language, specially chapter 5 (Pointers and arrays).
You can find a PDF copy easily. It's where I finally learned this concept.
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;
}
This is giving me a segfault at the memset and I have no idea why, I am going to a specific index of a 2D array, this should give me a char pointer and allow me to use memeset.
void test(char** test)
{
int i;
for(i=0;i<20;i++)
{
memset(test[i],0,sizeof(char)*1);
return;
}
}
int main()
{
char thing[20][20];
int i;
for(i=0;i<20;i++)
{
memset(thing[i],0,sizeof(char)*20);
}
test(thing);
return 0;
}
Your parameter declaration is incorrect, it should be:
void test(char test[20][20])
or:
void test(char test[][20])
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 have this code:
void PrintMainParameters(int n, char* array[])
{
int i = 0;
for(i = 0; i < n; i++)
{
printf("%s \n", array[i]);
}
}
int main(int argc, char* argv[] )
{
PrintMainParameters(argc, argv);
}
Works fine. Now I want to write PrintMainParameters as prototype to declare the function later in the source file.
I tried this one, but it says type mismatch, that the second parameter is an incompatible pointer type. I understand the compiler error, but I do not know why it occurs.
void PrintMainParameters(int, char*);
int main(int argc, char* argv[] )
{
PrintMainParameters(argc, argv);
}
void PrintMainParameters(int n, char* array[])
{
int i = 0;
for(i = 0; i < n; i++)
{
printf("%s \n", array[i]);
}
}
How must the prototype look like? Why does my code not work?
Your function takes an array of char pointers. Your prototype declares it to take a single char pointer instead. The correct prototype looks like this:
void PrintMainParameters(int, char*[]);
You can use either:
void PrintMainParameters(int, char**);
or:
void PrintMainParameters(int, char *[]);
Or if you prefer, you can insert a dummy parameter into the prototype, such as:
void PrintMainParameters(int argc, char *argv[]);