C - Error with mallocing a string - c

Question

I keep getting the following errors when i try to compile my code below. I searched online for what the errors meant but nothing seemed to fit my code. Please could someone tell me where im going wrong in malllocing a string below. I want to make 5 units in that array, and this is the first one.
Errors:
assignment4.c: In function ‘readFile’:
assignment4.c:42:19: warning: assignment makes integer from pointer without a cast [enabled by default]
assignment4.c:44:2: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat]
Code:
void readFile()
{
char * listofdetails;
listofdetails = malloc(sizeof(char)*40);
listofdetails[1] = "cars";
printf("%s \n", listofdetails[1]);
free(listofdetails);
}

Answer 1

Use strcpy (don't forget to #include <string.h>):
strcpy(listofdetails, "cars");
Also, "%s", expects a string, not a char, if you want to print a string
printf("%s \n", listofdetails);
is what you want.
If you want to print the first character in the string:
printf("%c \n", listofdetails[0]);
But if you want a list of details (as the name suggests), you need to reserve space for the list of n string (char **), and for each string (char *):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
extern char *strdup(const char *);
int main(void)
{
char **listofdetails = malloc(sizeof(*listofdetails) * 40);
if (listofdetails == NULL) {
perror("malloc");
exit(EXIT_FAILURE);
}
listofdetails[0] = strdup("cars");
if (listofdetails[0] == NULL) {
perror("strdup");
exit(EXIT_FAILURE);
}
printf("%s\n", listofdetails[0]);
free(listofdetails[0]);
free(listofdetails);
return 0;
}
Note that strdup is not part of the standard (but is available on many implementations).
Finally, if those strings are known at compile time you don't need to use malloc at all:
const char *listofdetails[] = {
"cars",
"trucks",
"motorcycles",
/* etc */
};

Answer 2

The two lines
char * listofdetails;
listofdetails = malloc(sizeof(char)*40);
can be written as
char * listofdetails = malloc(40);
(sizeof(char) is defined to be 1). Still you should not use magic numbers as 40 directly in the code but use aptly named constant
#define STRING_LENGTH 40
The line
listofdetails[1] = "cars";
is wrong, since the left side is the second character location in the list of details, and you try to assign a string character to it. Use
strcpy(listofdetails, "cars");
and the same problem in the line
printf("%s \n", listofdetails[1]);
Use
printf("%s \n", listofdetails);
instead.

Answer 3

listofdetails[1] is char(or int) type. %s expects a char * type. The fix:
printf("%s \n", listofdetails);
If you want to print out listofdetails[1], then use %c:
printf("%c \n", listofdetails[1]);
listofdetails[1] = "cars"; tries to assign a char array to a char(or int).
Instead, you can do:
listofdetails = malloc(sizeof(char)*40);
strcpy(listofdetails, "cars");

Answer 4

It would appear that you want a list of strings, but you're declaration of listofdetails is only a single string.
Instead, you need to add another pointer, like so:
char **listofdetails = malloc(sizeof(char *) * 40);
Now you can use listofdetails as you intend. Please note, however, that in C, array indexes start at 0, not 1. So you probably want to use listofdetails[0] = "cars"; to set the first element of your array.

Answer 5

You are assigning a string to a single byte location.
listofdetails[1] = "cars"; // Problem
where listofdetails[1] indicates a single character in a character array.
You should not assign string like above. use strcpy-
strcpy(listofdetails,"cars");
Else you can assign like this also
char *listofdetails = "cars";
It does not require memory allocation.
Try the below changes-
void readFile()
{
char * listofdetails;
listofdetails = malloc(sizeof(char)*40);
strcpy(listofdetails,"cars");
printf("%s \n", listofdetails);
free(listofdetails);
}
Use string.h and stdlib.h header file also.

Answer 6

a char is a single character (e.g. 'a')
char* is a pointer. in your case you allocate 40 (single) characters (malloc) and assign the address of the first to listofdetails. use strcpy, as Alter Mann wrote, to copy a string literal or any null-terminated sequence of chars to your listofdetails. to complete Alter Mann's example:
char * listofdetails;
listofdetails = malloc(sizeof(char)*40);
strcpy(listofdetails, "cars"); // make sure that the string on the right does not exceed 40 characters!
printf("%s\n", listofdetails); // will print "cars" on a single line
free(listofdetails);
example for array of strings:
char ** listofdetails;
listofdetails = malloc(sizeof(char*) * ARRAYSIZE);
listofdetails[0] = malloc(sizeof(char) * STRBUFFERSIZE);
strcpy(listofdetails[0], "cars");
printf("%s\n", listofdetails[0]);
free(listofdetails[0]); // delete string
free(listofdetails); // delete string-array

Answer 7

assignment4.c:42:19: warning: assignment makes integer from pointer without a cast [enabled by default]
comes from the line
listofdetails[1] = "cars";
listofdetails is an expression of type char *; listofdetails[1] is an expression of type char, which is an integral type, and it evaluates to the second element of the array that listofdetails points to. "cars" is an expression of type char * (more on that later), and you can't assign pointer values to integers without a cast.
This boils down to what listofdetails is supposed to store. Is it suposed to store a single string? If so, then you would need to write
strcpy( listofdetails, "cars" );
in order to copy the contents of the string literal "cars" to the memory listofdetails points to.
If it's supposed to store a pointer to an existing string or string literal, then you would skip the malloc step altogether and simply write
char *listofdetails = "cars";
The string literal "cars" is stored as a 5-element array of char such that it is allocated when the program starts up and released when the program exits. An array expression will be converted ("decay") to a pointer expression unless it is the operand of the unary & or sizeof operators, or is a string literal being used to initialize another array in a declaration.
In the case above, "cars" isn't the operand of the sizeof or unary & operators, and it isn't being used to initialize an array of char, so it decays to a char *, and its value is the address of the first character in the string. So by the time this is done, listofdetails points to the string literal "cars".
If it's supposed to store a list of pointers to strings, then you would need to do the following:
char **listofdetails = malloc( sizeof *listofdetails * 40 ); // note extra level of indirection
This will allocate enough memory to store 40 pointers to char; since the type of listofdetails is char **, the type of listofdetails[1] will be char *, so the following lines will work as you intend:
listofdetails[1] = "cars";
printf("%s \n", listofdetails[i] );
Making that last change will fix
assignment4.c:44:2: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat]
which is complaining about
printf("%s \n", listofdetails[1] );
since as currently written, listofdetails[1] has type char, which is promoted to int since printf is a variadic function, and %s expects the corresponding argument to have type char *.
One more thing; in C, arrays are 0-origin, so you'd use the index 0 to access the first element of the list, and index 39 to access the last one.