int main() {
int i;
char a[]={"Hello"};
while(a!='\0') {
printf("%c",*a);
a++;
}
getch();
return 0;
}
Strings are stored in contiguous memory locations & while passing the address to printf() it should print the character. I have jst started learning C. I am not able to find an answer to this. Pls help.
Well a is the name of the array which you cannot increment. It is illegal to change the address of the array.
So define a pointer to a and then increment
#include <stdio.h>
#include <conio.h>
int main()
{
int i;
char a[]="Hello";
char *ptr = a;
while(*ptr!='\0')
{
printf("%c",*ptr);
// a++ here would be illegal
ptr++;
}
getch();
return 0;
}
NOTE:
In fact, arrays in C are non-modifiable lvalues. There are no
operations in C that can modify the array itself (only individual
elements can be modifiable).
In your code, a is a name of an array, you can't modify it like a++. Use a pointer like this:
char *p = "Hello";
while(*p++)
{
printf("%c",*p);
}
Three problems:
char a[]={"Hello"}; is illegal. {"Hello"} can only initialize char* a[]. You probably want char a[]="Hello";
while(a!='\0') - you probably meant *a != '\0'. a is the array itself.
a++; - an array cannot be incremented. you should increment a pointer pointing to it.
You can also try it using a for loop:
#include <stdio.h>
int main(void) {
char a[] = "Hello";
char *p;
for(p = a; *p != '\0'; p++) {
printf("%c", *p);
}
return 0;
}
Related
I have been trying to solve this issue for whole day, and could not do it on my own. Searching the internet didn't help me solve it either
So, this the function prototype:
void invert(char **arr, int n);
First argument is an array of strings, and the second one is number of strings in an array.
This is my code:
#include <stdio.h>
#include <string.h>
void invert(char** arr, int n)
{
int i, j, len;
for(j=0;j<n;j++)
{
len=strlen(arr[j]);
for(i=0;i<len/2;i++)
{
char tmp = arr[j][i];
arr[j][i] = arr[j][len - i - 1];
arr[j][len - i - 1] = tmp;
}
}
}
int main()
{
int n=3, i;
char **arr;
arr[0]="John";
arr[1]="Doe";
arr[2]="Programmer";
invert(arr, n);
for(i=0;i<3;i++)
{
printf("%s ",arr[i]);
}
}
The code breaks when it reaches the line:
arr[j][i] = arr[j][len - i - 1];
and I can't figure out why.
The function receives an array of strings perfectly (tested it with some printf statements for characters of specific strings), and the char tmp succesfully recieves a correct character, but the program crashed when it reaches the line mentioned earlier. Printf statements after that line don't work.
Did I miss anything? Can someone explain what am I doing wrong? Thank you!
For starters this code snippet
char **arr;
arr[0]="John";
arr[1]="Doe";
arr[2]="Programmer";
invokes undefined behavior because the pointer arr is uninitialized and has an indeterminate value.
Moreover this approach in any case is wrong because you may not change string literals.
What you need is to declare a two-dimensional array as for example
enum { N = 11 };
//...
char arr[3][N] =
{
"John", "Doe", "Programmer"
};
In this case the function declaration will look like
void invert( char arr[][N], int n );
The enumeration must be declared before the function declaration.
Instead of the two-dimensional array you could declare an array of pointers like
char s1[] = "John";
char s2[] = "Doe";
char s3[] = "Programmer";
char * arr[3] = { s1, s2, s3 };
In this case the function declaration may be as shown in your question
void invert(char** arr, int n)
So what you need to do with minimal changes is to substitute this code snippet
char **arr;
arr[0]="John";
arr[1]="Doe";
arr[2]="Programmer";
for this code snippet
char s1[] = "John";
char s2[] = "Doe";
char s3[] = "Programmer";
char * arr[3] = { s1, s2, s3 };
To begin with, what you have here:
char **arr;
is a pointer to pointer to char.
Secondly, even if you had an array of pointers to char, like so :
char *arr[3];
And then assigning each string literal :
arr[0]="John";
arr[1]="Doe";
arr[2]="Programmer";
would still invoke Undefined behavior, since you are attempting to modify a string literal which is read only.
What you need is, either a 2D array of chars :
char arr[][100] = {"John", "Doe", "Programmer"};
and also change the function signature to :
void invert(char arr[][100], int n)
or you have to dynamically allocate memory and use a function like strcpy(), strdup(), memcpy() etc :
char **arr;
arr = malloc(n * sizeof(char *)); // or sizeof(*arr)
if (arr == NULL) {
fprintf(stderr, "Malloc failed to allocate memory\n");
exit(1);
}
arr[0] = strdup("John"); // good idea to also check if strdup returned null
arr[1] = strdup("Doe");
arr[2] = strdup("Programmer");
invert(arr, n);
for(i=0;i<3;i++)
{
printf("%s ",arr[i]);
}
for (i = 0; i < 3; i++) {
free(arr[i]);
}
free(arr);
This relates to C. I am having some trouble understanding how I can assign strings to char pointers within arrays from a function.
#include <stdio.h>
#include <string.h>
void function(char* array[]);
int main(void)
{
char* array[50];
function(array);
printf("array string 0: %s\n",array[0]);
printf("array string 1: %s\n",array[1]);
}
void function(char* array[])
{
char temp[] = "hello";
array[0] = temp;
array[1] = temp;
return;
}
Ideally, I would like the main printf function to return
array string 0: hello
array string 1: hello
But I'm having trouble understanding arrays of pointers, how these pass to functions and how to manipulate them in the function. If I declare a string like char temp[] = "string" then how do I assign this to one of the main function array[i] pointers? (assuming I have my jargon right)
char temp[] = "hello"; only creates a local, temporary array inside the function. So when the function exists, the array will be destroyed.
But with array[0] = temp; you're making array[0] point to the local array temp.
After the function returns, temp doesn't exist anymore. So accessing array[0] which pointed to temp will cause undefined behavior.
You could simply make temp static, so it also exists outside the function:
static char temp[] = "hello";
Or, you could copy the "hello" string to array[0] and array[1]. For copying C-strings, you normally use strcpy.
char temp[] = "hello";
strcpy(array[0], temp);
strcpy(array[1], temp);
However, before copying you need to make sure array[0] and array[1] point to memory that has enough space to hold all characters of "hello", including the terminating null character. So you have to do something like this before calling strcpy:
array[0] = malloc(6);
array[1] = malloc(6);
(6 is the minimum numbers of characters that can hold "hello".)
how do I assign this to one of the main function array[i] pointers
Arrays cannot be assigned.
A pointer cannot hold an array, it can only refer to an array. For the latter the pointer needs to get an array's address assigned.
Referring 1.
This
char temp[] = "hello";
isn't an assigment, but an initialisation.
This
char temp[];
temp[] = "hello";
would not compile (the 2nd line errors), as an array cannot be assigned.
Referring 2.
This
char* array[50];
defines an array of 50 pointers to char, it could reference 50 char-arrays, that is 50 C-"strings". It cannot hold the C-"strings" themselfs.
Example
Applying what is mentioned above to your code whould lead to for example the following:
#include <stdio.h>
#include <string.h>
void function(char* array[]);
int main(void)
{
char* array[50];
/* Make the 1st two elements point to valid memory. */
array[0] = malloc(42); /* Real code shall test the outcome of malloc()
as it might fail and very well return NULL! */
array[1] = malloc(42);
function(array);
printf("array string 0: %s\n",array[0]);
printf("array string 1: %s\n",array[1]);
return 0;
}
void function(char* array[])
{
char temp[] = "hello";
/* Copy the content of temp into the memory that was allocated to
the array's 1st memebrs in main(). */
strcpy(array[0], temp);
strcpy(array[1], temp);
return;
}
first, you need to allocate the destination.
second, char temp[] = "hello"; in function() is local variable. you cannot use their outside of the function. use strcpy to copy the content.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void function(char* dest)
{
char temp[] = "hello";
strcpy(dest, temp);
return;
}
int main(void)
{
// or just char dest[10] = {0};
char *dest = malloc(10);
function(dest);
printf("dest: %s\n", dest);
}
In you program, you defined char* array[50];, so you need to create memory space for each item:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void function(char* a[])
{
char temp[] = "hello";
strcpy(a[0], temp);
strcpy(a[1], temp);
return;
}
int main(void)
{
char *a[50];
int i = 0;
for (i = 0; i < 50; ++i)
a[i] = malloc(10);
function(a);
printf("a[0]: %s\n", a[0]);
printf("a[1]: %s\n", a[1]);
}
I have a char *p = "abcd", how can I access the elements 'a','b','c','d' using only C(not C++)? Any help would be appreciated .
You can use indexing:
char a = p[0];
char b = p[1];
/* and so on */
Equivalently you can use pointer arithmetic, but I find it less readable:
char a = *p;
char b = *(p+1);
If you really want to surprise someone you can also write this:
char a = 0[p];
char b = 1[p];
/* and so on */
Here, p refers an array of character pointer. You ca use the array indexing to access each variable in that array. The most widely used notation for this is p[n], where n is the n+1th character [element].
example:
for the 1st character, use p[0], 2nd character, use p[1] and so on..
another example:
#include <stdio.h>
int main()
{
char *p="abcd";
for (; *p; p++)
printf("%c\n", *p);
return 0;
}
result is:
a
b
c
d
Use the array subscript operator []. It allows you to access the nth element of a pointer type in the form of p[n].
You can also increment the pointer by using the increment operator ++.
#include <stdio.h>
int main()
{
char *p="abcd";
printf("%c\n", p[0]);
printf("%c\n", p[1]);
printf("%c\n", p[2]);
printf("%c\n", p[3]);
return 0;
}
returns
a
b
c
d
#include <stdio.h>
#include <stdlib.h>
void main()
{
char* str="Hello, World!";
printf("%s\n",str);
}
Output:
Hello, World!
2nd Method:
#include <stdio.h>
#include <stdlib.h>
void main()
{
char* str="Hello";
int n=strlen(str);
for(int i=0;i<n;i++)
{
printf("%c"str[i]);
}
}
Output:
Hello
I am trying to write a function, uppercase, that converts all lowercase characters in a string into their uppercase equivalents.
However, I am getting a Bus 10 error in my code. I know that string literals cannot be modified in C; so, I am not sure if this is the right approach.
My code is below:
#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>
#include <string.h>
int uppercase(char source[])
{
int i;
for(i=0; i<=strlen(source); ++i)
if (source[i]>= 'a' && source[i]<= 'z')
source[i]= source[i]-'a' +'A';
else
source[i]=source[i];
}
int main(){
uppercase("cold");
return 0;
}
Ideally this function should return COLD.I suppose the error lies in my whole if statement.
The reason you get a crash is that your code modifies a string literal. Characters inside string literals are placed in protected memory area, and therefore may not be changed: it us undefined behavior.
Replace this
uppercase("cold");
with this:
char cold[] = "cold";
uppercase(cold);
Now the characters of the string are placed in a modifiable area of memory, allowing you to make changes as needed.
Your absolutly working with pointers without even to know it.
In your function definition
int uppercase(char source[])
char source[] is considered by the compiler as a pointer to char (char *source)
So when passing a string literal to uppercase() your just passing it's adress. Then in your function your trying to modify it which leads to undefined behaviour.
Also you can't return a whole array so you just return a pointer to it.
char *uppercase(char source[])
{
int i;
size_t len = strlen(source);
char *tmp;
tmp = malloc(len+1);
if (tmp!=NULL){
memcpy(tmp, source, len+1);
for(i=0; i<len; ++i){
if (tmp[i]>= 'a' && tmp[i]<= 'z'){
tmp[i]= tmp[i]-'a' +'A';
}
}
}
return tmp;
}
Then:
int main(){
char *str = uppercase("cold");
printf("%s", str);
free(str);
return 0;
}
You complete code: http://ideone.com/BJHDIF
I need help figuring out why I am getting a segmentation fault here. I have gone over it and I think I am doing something wrong with the pointers, but I can figure out what.
My Program:
#include <stdlib.h>
#include <stdio.h>
void encrypt(char* c);
//characters are shifted by 175
int main(){
char* a;
*a = 'a';
/*SEGMENTATION FAULT HERE!*/
encrypt(a);
printf("test:%c/n",*a);
return 0;
};
void encrypt(char* c){
char* result;
int i=(int)(*c);
i+=175;
if(i>255)
{
i-=256;
}
*c=(char)i;
};
The problem is here:
char *a;
*a = 'a'
Since the variable "a" is not initialized, *a = 'a' is assigning to a random memory location.
You could do something like this:
char a[1];
a[0] = 'a';
encrypt(&a[0]);
Or even just use a single character in your case:
int main(){
char a = 'a';
encrypt(&a);
printf("test:%c/n",a);
return 0;
};
char* a;
*a = 'a';
/*SEGMENTATION FAULT HERE!*/
There isn't any "there" there. You've declared a and left it uninitialized. Then you tried to use it as an address. You need to make a point to something.
One example:
char buffer[512];
char *a = buffer;
(Note buffer has a maximum size and when it falls out of scope you cannot reference any pointers to it.)
Or dynamic memory:
char *a = malloc(/* Some size... */);
if (!a) { /* TODO: handle memory allocation failure */ }
// todo - do something with a.
free(a);
That pointer a does not get the actual space where to store the data. You just declare the pointer, but pointing to where? You can assign memory this way:
char *a = malloc(1);
Then it won't segfault. You have to free the variable afterwards:
free(a);
But in this case, even better,
char a = 'a';
encript(&a);