I expected to get errors in following code, but I did not. I did not use & sign. Also I am editing array of chars.
#include <stdio.h>
int main()
{
char name[10] ="yasser";
printf("%s\n",name);
// there is no error ,
// trying to edit array of chars,
// also did not use & sign.
scanf("%s",name);
// did not use strcpy function also.
printf("%s\n",name);
return 0;
}
I expected to get errors in following code, but I did not.I did not use & sign.
scanf("%s",name);
That's totally ok as name is already the address of the character array.
It sounds like you have several questions:
calling scanf("%s", name) should have given an error, since %s expects a pointer and name is an array? But as others have explained, when you use an array in an expression like this, what you always get (automatically) is a pointer to the array's first element, just as if you had written scanf("%s", &name[0]).
Having scanf write into name should have given an error, since name was initialized with a string constant? Well, that's how it was initialized, but name really is an array, so you're free to write to it (as long as you don't write more than 10 characters into it, of course). See more on this below.
Characters got copied around, even though you didn't call strcpy? No real surprise, there. Again, scanf just wrote into your array.
Let's take a slightly closer look at what you did write, and what you didn't write.
When you declare and initialize an array of char, it's completely different than when you declare and initialize a pointer to char. When you wrote
char name[10] = "yasser";
what the compiler did for you was sort of as if you had written
char name[10];
strcpy(name, "yasser");
That is, the compiler arranges to initialize the contents of the array with the characters from the string constant, but what you get is an ordinary, writable array (not an unwritable, constant string constant).
If, on the other hand, you had written
char *namep = "yasser";
scanf("%s", namep);
you would have gotten the problems you expected. In this case, namep is a pointer, not an array. It's initialized to point to the string constant "yasser", which is not writable. When scanf tried to write to this memory, you probably would have gotten an error.
When you pass arrays to functions in C, they decay to pointers to the first item.
Therefore for:
char name[] ="yasser";
scanf("%s", name) is the same as scanf("%s", &name[0]) and either of those invocations should send shivers down your spine, because unless you control what's on your stdin (which you usually don't), you're reading a potentially very long string into a limited buffer, which is a segmentation fault waiting to happen (or worse, undefined behavior).
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char **argv, char **envp) {
char *myName = (char *) calloc(10, sizeof(char));
*(myName)='K'; *(myName+1)='h'; *(myName+2)='a'; *(myName+3)='l'; *(myName+4)='i'; *(myName+5)='d';
printf("%s\n",myName);
scanf("%s",myName);
printf("%s\n",myName);
return (EXIT_SUCCESS);
}
#include <stdio.h>
#include <string.h>
int main()//fonction principale
{
char name[10] ="yasser";
int longeur=0;
printf("%s\n",name);
scanf("%s",name);
longeur = strlen(name);
for (int i=0;i<longeur;i++) {
printf("%c",*(name+i));
}
return 0;}
Related
I am trying to write code to implement strchr function in c. But, I'm not able to return the string.
I have seen discussions on how to return string but I'm not getting desired output
const char* stchr(const char *,char);
int main()
{
char *string[50],*p;
char ch;
printf("Enter a sentence\n");
gets(string);
printf("Enter the character from which sentence should be printed\n");
scanf("%c",&ch);
p=stchr(string,ch);
printf("\nThe sentence from %c is %s",ch,p);
}
const char* stchr(const char *string,char ch)
{
int i=0,count=0;
while(string[i]!='\0'&&count==0)
{
if(string[i++]==ch)
count++;
}
if(count!=0)
{
char *temp[50];
int size=(strlen(string)-i+1);
strncpy(temp,string+i-1,size);
temp[strlen(temp)+1]='\0';
printf("%s",temp);
return (char*)temp;
}
else
return 0;
}
I should get the output similar to strchr function but output is as follows
Enter a sentence
i love cooking
Enter the character from which sentence should be printed
l
The sentence from l is (null)
There are basically only two real errors in your code, plus one line that, IMHO, should certainly be changed. Here are the errors, with the solutions:
(1) As noted in the comments, the line:
char *string[50],*p;
is declaring string as an array of 50 character pointers, whereas you just want an array of 50 characters. Use this, instead:
char string[50], *p;
(2) There are two problems with the line:
char *temp[50];
First, as noted in (1), your are declaring an array of character pointers, not an array of characters. Second, as this is a locally-defined ('automatic') variable, it will be deleted when the function exits, so your p variable in main will point to some memory that has been deleted. To fix this, you can declare the (local) variable as static, which means it will remain fixed in memory (but see the added footnote on the use of static variables):
static char temp[50];
Lastly, again as mentioned in the comments, you should not be using the gets function, as this is now obsolete (although some compilers still support it). Instead, you should use the fgets function, and use stdin as the 'source file':
fgets(string, 49, stdin);/// gets() has been removed! Here, 2nd argument is max length.
Another minor issue is your use of the strlen and strncpy functions. The former actually returns a value of type size_t (always an unsigned integral type) not int (always signed); the latter uses such a size_t type as its final argument. So, you should have this line, instead of what you currently have:
size_t size = (strlen(string) - i + 1);
Feel free to ask for further clarification and/or explanation.
EDIT: Potential Problem when using the static Solution
As noted in the comments by Basya, the use of static data can cause issues that can be hard to track down when developing programs that have multiple threads: if two different threads try to access the data at the same time, you will get (at best) a "data race" and, more likely, difficult-to-trace unexpected behaviour. A better way, in such circumstances, is to dynamically allocate memory for the variable from the "heap," using the standard malloc function (defined in <stdlib.h> - be sure to #include this header):
char* temp = malloc(50);
If you use this approach, be sure to release the memory when you're done with it, using the free() function. In your example, this would be at the end of main:
free(p);
I have a char pointer: char *sentences; and I read it like this:
#include <stdio.h>
int main() {
char *sentences;
sentences="We test coders. Give us a try?";
printf("%s", sentences);
return 0;
}
but I want to read with scanf() function in c.
scanf("%[^\n]s",S); or scanf("%s",S); didn't work.
How can I do that?
Are you declaring the variable char *sentences; and immediately trying to write to it with scanf? That's not going to work.
Unless a char * pointer is pointing to an existing string, or to memory allocated with malloc-family functions, assigning to it with scanf or similar is undefined behavior:
char *sentences;
scanf("%s", sentences); // sentences is a dangling pointer - UB
Since you haven't actually shared your code that uses scanf and doesn't work, I can only assume that's the problem.
If you want to assign a user-supplied value to a string, what you can do is declare it as an array of fixed length and then read it with a suitable input function. scanf will work if used correctly, but fgets is simpler:
char sentence[200];
fgets(sentence, 200, stdin);
// (User inputs "We test coders. Give us a try")
printf("%s", sentence);
// Output: We test coders. Give us a try.
Also, never, ever use gets.
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I have this code:
char** SplitToWords(char* str);
int main()
{
char** wordarr;
char str[] = "This is a sentence";
wordarr = SplitToWords(str);
return 0;
}
After the main comes the function implementation.
I am not sure the following does what I want it to do (i.e. receive an array of strings from a function):
wordarr = SplitToWords(str);
I somehow managed to convince the compiler that it's ok, but I assume it just does something else.
If it does, how do I find out the length of the array (the number of strings in it).
Thanks
I'll try to quickly visit all aspects you might not yet fully understand:
A string in C is described as a contiguous sequence of chars, ending with a char of value 0 (as a literal: '\0'). It is not a first class object, therefore hasn't its own type. So what you use to hold a string is an array of char. Therefore, taking your question by the word, "receive an array of strings from a function" is not possible.
An array is a contiguous sequence of objects of the same type. In C, the identifier of an array doesn't have a value itself; when it's evaluated, it decays as a pointer to the array's first element instead. This is especially important when passing arrays to functions or returning them from functions -- you can't actually pass the array, you always pass a pointer
e.g. you could write:
char x[] = "foo"; // initialize a char array from a string literal
char *xp = x; // here, x evaluates as a pointer to the first element of the array
You already use pointer types for your function's argument and return value, I just think it's quite important to understand what happens entirely.
You write char** SplitToWords(char* str); and ask whether this returns an "array of strings" -- well, sort of, as you should understand after reading 1. and 2. -- What it does is returning a pointer to char *. This pointer could be a pointer to the first element of an array. So in this case, it would return a pointer to an array of char * pointers. Each of these pointers could itself be a pointer to an array of chars, therefore point to a string. But what's very important is to understand you never return an array, you always return a pointer to it. It's so important because:
You might get the idea to do something like this:
char** SplitToWords(char* str)
{
char *words[16];
// code to fill `words` with pointers to the actual words
return words; // WRONG!
}
Here, because you're not returning the array words but a pointer to it (see point 2), you return a pointer to an object that no longer exists. words is in the scope of your function and has automatic storage duration, that means it only lives as long as the execution is inside of the function. One solution would be to declare words with the static storage class specifier. This way, it lives for the entire execution time of the program. But be aware that this also means there's only a single instance ever, it's always the same object. This will be a major headache for threaded programs, for example. The other way around is to dynamically allocate words using malloc(). But then, the caller of the function must free() it later.
As for your second question, how to let the caller know the number of words -- it's in the comments already, but just for completeness, a typical approach to solve this is to append another entry that is a NULL pointer. So the caller can iterate over the pointers until it finds NULL.
Regarding your comment, of course you can create the array outside the function and pass a pointer to the function, so the function only fills it. This is a common idiom in C (e.g. think about fgets(), which takes a pointer to the char array that's filled with a string by the function).
Functions working this way will need an additional size_t parameter, so they know the size of the array they should fill through the pointer, otherwise you'd have the risk of buffer overflows (this is why gets() was finally removed from the C standard). If you decide that the caller provides the storage, your function should have this prototype:
// returns the number of words found, up to `nwords`
size_t SplitToTwords(char **words, size_t nwords, char *str);
It should be called e.g. like this:
char *words[16];
size_t nwords = SplitToWords(words, 16, "the quick brown fox"); // returns 4
Remember that the strings holding the words themselves need storage as well. You can either manipulate the bytes in str to insert a '\0' after each word, overwriting the first whitespace character (this is what strtok() does) or you can copy the words to new strings, but then you would have to malloc() each of them again and the caller has to free() them later.
Yes, you could solve it by using a function with return value char **. However, there's no way to find out how many words there are afterwards.
You can solve this by allocating one more element for the return pointer and set it to NULL. Then you can get the number of words with this code:
wordarr = SplitToWords(str);
char **ptr=wordarr;
int noWords=0;
while(!*(ptr+noWords))
noWords++;
But if you want to return multiple data in C, you either need to define a return struct or using return arguments. In this case, it could look like this for the first option:
typedef struct wordList {
char **wordarr;
int noWords;
}
wordList SplitToWords(char* str);
And the second:
char** SplitToWords(char* str, int *noWords);
or
void SplitToWords(char* str, char*** wordarr, int *noWords);
Note that there's three *. That's because we want it to be a pointer to char **
#include "stdafx.h"
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAXSTRINGS 5000
int main(int argc, char *argv[]) {
char *stringTable[MAXSTRINGS];
char sentence[] = "This is a sentence";
char *token = NULL;
int i = 0;
while ((token = strtok(token == NULL ? sentence : NULL, " ")) != NULL)
{
printf("%s\n\r", token);
stringTable[i] = (char *)malloc(strlen(token) + 1); //have no "plain" C compiler - VS C++ used so cast needed :)
strcpy(stringTable[i++], token);
}
stringTable[i] = NULL; // if you need to iterate through later
printf("%d tokens found\n\r", i);
for (int y = 0; y < i; y++)
free(stringTable[y]);
}
Here's more code I whipped up since i was having trouble with my major program that I now fixed.
I have a function which modifies a series of bytes. In this example, the function is supposed to fill up the first 9 bytes of the char array with the numbers 1 through 9 consecutively.
Two tests are run. The first one is calling the function where the parameter is (char*)&myvar. The second test only uses myvar as a parameter. I thought I always had to use an & in front of a char array pointer when I want the string returned in the parameter portion of the function.
Why does this program only work when I don't prepend (char*)& to my char array variable?
When I do apply it, I receive a segmentation fault.
#include "stdio.h"
#include "stdlib.h"
#include "string.h"
int func(char* abc){
char *p=abc;
int n;
for (n=1;n<10;n++){
*p=(unsigned char)(n+48);p++;
}
return 0;
}
int main(){
char vars[1000]="234";
char *myvar=vars;
printf("Function test\n");
int result=func((char*)&myvar); //causes segfault
printf("Function result %s\n",myvar); //segfault here
printf("Function test again\n");
int result2=func(myvar); //works
printf("Function result %s\n",myvar);
printf("DONE\n");
return 0;
}
Why does this program only work when I don't prepend (char*)& to my char array variable?
Because doing that is completely wrong and not a thing that makes sense.
I thought I always had to use an & in front of a char array pointer when I want the string returned in the parameter portion of the function.
You don't. (Also, you do not have a "char array pointer", and "when I want the string returned in the parameter portion of the function" doesn't make sense.)
When you need to pass a char * to a function that takes a char *, you do not need to put any special prefix in front of the pointer. You just pass it directly, the way you did with
int result2=func(myvar);
You could also have passed in vars, due to the automatic conversion from an array to a pointer to its first element, just like you were able to do char *myvar=vars; without any special casting.
#include <stdio.h>
#include <string.h>
int main(void) {
int number_of_members;
char family[number_of_members][20][number_of_members][20];
char member_name[20];
char birth_state[20];
char family_last_name[20];
printf("What is the last name of the family?\n");
scanf("%s", &family_last_name);
printf("How many members do you want to create?\n");
scanf("%d", &number_of_members);
int const FAMILY_SIZE = number_of_members;
number_of_members = number_of_members -1;
printf("Enter the family member name: \n");
for(number_of_members;number_of_members>-1;number_of_members--)
{
scanf("%s", &member_name);
strcpy(family[number_of_members], member_name);
printf(" %d %s %s\n",number_of_members, member_name, family_last_name);
}
printf("%s, %s ", family[0], family[1]);
return 0;
}
Here is the output:(from Ideone.com)
Ideone.com with code
The input to this code is: Layne , 2 , tim , jim.
When run, it shows the correct index with the name in the array however, once out it will show the last entered name, jim, as family1 and family[0]. Am I not understanding how strcpy() works? or is it a logic error?Some assistance soon would be appreciated!
This is very very wrong
int number_of_members;
char family[number_of_members][20][number_of_members][20];
Because you haven't initialized number_of_members.
Because it doesn't make sense whatsoever, it's not possible that you really need this kind of array.
And yes, if you enable compiler warnings it will hit you in your nose with a stick, because
strcpy(family[number_of_members], member_name);
shouldn't even compile and is undefined behavior since the type of family[number_of_members], is an array of arrays of arrays of char.
strcpy can take an array of char's because it will be automatically converted to a char poitner, and provided that the contents of the array comply with what a c string is, then strcpy() will work correctly, in your case the behavior is undefined because almost surely the '\0' will never be found in the destination pointer.
instead of
int num_of_members;
char family[number_of_members][20][number_of_members][20];
which is not C code, do this
#define MAX_MEMBERS 20
char family[MAX_MEMBERS][20];
which creates a rectangular array of arrays each of 20 bytes long