#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(){
int temptn;
char *ticno, temptn2[10];
ticno = (char*)malloc(100);
printf("%s", ticno);
for (int i=0; i < 6; i++){
temptn = i;
sprintf(temptn2, "%d", temptn);
strcat(ticno, temptn2);
}
printf("\n%s", ticno);
// prints &┐012345
}
Does anyone know how to solve the problem of getting unknown characters in the string after using malloc on the char *ticno?
malloc() just allocates memory, it doesn't zero it for you, so there's likely to be arbitrary garbage in there.
You can do
ticno = (char*)malloc(100);
memset(ticno, 0, 100);
to zero those 100 bytes, or use calloc, which does zero the memory:
ticno = (char*)calloc(100, 1);
Related
I have a question regarding the difference between char *<variable>[6] and char (*<variable>)[6].
I was making a code for an exercise which asks to input and store 6 strings and later turn each into an integer. My plan was to make an array of strings to store them and I did char *<variable>[6] = malloc(7*sizeof(char)); but an error showed up saying ' array initializer must be an initializer list ' but when i changed it to char (*<variable>)[6] = malloc(7*sizeof(char)); it can work properly. What is the difference between (*<variable>)[6] and *<variable>[6] ?
My code below:
#include <stdio.h>
#include <stdlib.h>
int main(void){
char (*sixString)m[6] = (char *) malloc(7 * sizeof(char));
int i;
int sum = 0;
for(i = 0; i < 6; i++){
printf("Input for the Number %d: ", i + 1);
scanf("%s", sixString[i]);
//fgets(sixString[i], 100, stdin);
printf("%s\n", sixString[i]);
sum = sum + atoi(sixString[i]);
}
float average = (float)sum/6 ;
printf("Sum is: %d\nAverage is: %.2f\n", sum, average);
free(sixString);
return 0;
}
char *<variable>[6]
is a 6-element array of char*.
char (*<variable>)[6]
is a pointer to 6-element array of char.
Your code allocated insufficient number of bytes. You should use
char (*sixString)[6] = malloc(7 * sizeof(char[6]));
or
char (*sixString)[6] = malloc(7 * sizeof(*sixString));
instead of
char (*sixString)m[6] = (char *) malloc(7 * sizeof(char));
(size is fixed and the extra m is removed)
Note that then you allocate 7 elements of char[6], but you use only 6 of them in the loop.
Allocating extra elements may look inefficient, but cause less harm than allocating insufficient size.
How do you make 2 array strings into 1 array string, where I can print out all the 52 playing cards?
my code:
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<string.h>
int main() {
char deck[52];
char suits[] = {"Hearts","Diamonds","Clubs","Spades"};
char values[]= {"Ace","Two","Three","Four","Five","Six",\
"Seven","Eight","Nine","Ten","Jack",\
"Queen","King"};
int V, S, d = 0;
char string;
for ( S= 0; S <4; S++) {
for (V =0; V< 13; V++) {
string = strcat( values[V], suits[S]);
deck[d] = string;
printf("%s\n", string);//prints out all the 52 playing cards
d++;
}
}
return 0;
}
When I executed the program, the problem comes up which asks me to debug the program or close the program, where I closed the program in the end, which returns nothing. Can you please give me the answer which works?
Check the below code which fixes the issues in your code:
The problem with your code is you try to modify the actual string before printing and because of this there is a modified string in the next iteration. So just copy the values and suits to array and print it out as shown below.
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<string.h>
int main()
{
int i=0;
char deck[30] = "";
char suits[][30] = {"Hearts","Diamonds","Clubs","Spades"};
char values[][30]= {"Ace","Two","Three","Four","Five","Six",
"Seven","Eight","Nine","Ten","Jack",
"Queen","King"};
int V, S;
for ( S= 0; S <13; S++)
{
for (V =0; V< 4; V++){
memset(deck,0,sizeof(deck));/* Clear the buffer before writing new value*/
strcpy( deck, values[S]);
strcat(deck,suits[V]);
printf("%s\n", deck);//prints out all the 52 playing cards
i++;
}
}
printf("Number of playing cards: %d\n",i);
return 0;
}
strcat() returns a char *, a pointer to a char, not a char.
You are not even required to even consider the return value of strcat() since the destination pointer (first argument) will now contain the concatenated string, assuming enough memory is already allocated.
So here in your code, you are trying to put the concatenated string to values[V] which could fail when memory already allocated to it becomes insufficient.
The best method would be to allocate some memory (as you did with deck[]) and set it all to zeroes. Then keep strcat()ing there.
strcat(deck, values[V]);
strcat(deck, suits[S]);
An alternative to using strcpy and strcat is to use sprintf.
#include<stdio.h>
#include<string.h>
#define NUM_SUITS 4
#define CARDS_PER_SUIT 13
#define TOTAL_CARDS (NUM_SUITS * CARDS_PER_SUIT)
int main()
{
char deck[TOTAL_CARDS][24];
char* suits[NUM_SUITS] = {"Hearts","Diamonds","Clubs","Spades"};
char* values[CARDS_PER_SUIT]= {"Ace","Two","Three","Four","Five","Six",
"Seven","Eight","Nine","Ten","Jack",
"Queen","King"};
int s, c, i;
for(s = 0; s < NUM_SUITS; s++)
{
for(c = 0; c < CARDS_PER_SUIT; c++)
{
sprintf(deck[(s * CARDS_PER_SUIT) + c], "%s of %s", values[c], suits[s]);
}
}
for(i = 0; i < TOTAL_CARDS; i++)
{
printf("%s\n", deck[i]);
}
return 0;
}
I just want you to ask what did I do wrong with this code.
I wrote a function that take a char* in parameter, I want to modify it directly without returning smthg, and reverse the string.
#include <iostream>
void reverseString(char *p_string){
int length = strlen(p_string);
int r_it = length - 1;
char* tmp = (char*)malloc(length);
int last_it = 0;
for (int i = 0; i != length; i++){
tmp[i] = p_string[r_it];
r_it--;
last_it++;
}
tmp[last_it] = '\0';
strcpy_s(p_string, length + 1, tmp);
//free(tmp);
}
int main(){
char str[] = "StackOverflow";
reverseString(str);
std::cout << str << std::endl;
system("pause");
}
I'm used to C++ and don't often use C functions like malloc/free/strcpy...
Here, my problem is, when I alloc memory for my temporary char, I called mallec(length) for length = 13 in this case, char = 1 bytes so it should be allocate memory for 13 char is that right?
Problem is allocate more space than need so i need to use '\0' before my strcpy_s if not it breaks.
Did I do a mistake somewhere?
Also, when i call free(tmp), it breaks too and say heap corruption, but I didn't free the memory before that.
Thanks for helping !
I took your original code and added a simple '+1' to the size of the malloc and got a passing result.
Not sure if your exercise is related specifically to the use of malloc, but have you considered doing the reversal directly inside the original string?
For example:
void reverseString(char *p_string){
char* p_end = p_string+strlen(p_string)-1;
char t;
while (p_end > p_string)
{
t = *p_end;
*p_end-- = *p_string;
*p_string++ = t;
}
}
int main(){
char str[] = "StackOverflow";
reverseString(str);
std::cout << str << std::endl;
system("pause");
}
If you are required to use malloc, then you need to ensure that you allocate enough space for string which includes the '\0'
You must use
int length = strlen(p_string);
int r_it = length - 1;
char* tmp = (char*)malloc(length+1);
Since strlen doesn't count the \0 character. So this will fail if you don't use length+1:
tmp[last_it] = '\0';
The length of a C string is determined by the terminating
null-character: A C string is as long as the number of characters
between the beginning of the string and the terminating null character
(without including the terminating null character itself).
http://www.cplusplus.com/reference/cstring/strlen/
Btw. C99 support semi dynamic arrays. So could you try this:
char tmp[length+1];
Source:
http://en.wikipedia.org/wiki/Variable-length_array
float read_and_process(int n)
{
float vals[n];
for (int i = 0; i < n; i++)
vals[i] = read_val();
return process(vals, n);
}
Check the below C code:
The memory allocated to tmp should be length+1 as done below and also there are many unnecessary variables which can be avoided.
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
void reverseString(char *p_string){
int i;
int length = strlen(p_string);
int r_it = length - 1;
char* tmp = (char*)malloc(length+1);
for (i = 0; i != length; i++){
tmp[i] = p_string[r_it--];
}
tmp[i] = '\0';
strcpy(p_string, tmp);
return;
}
int main(){
char str[] = "StackOverflow";
reverseString(str);
printf("%s",str);
return 0;
}
There is nothing fundamentally wrong with your approach, just some of the details. Since I am not sure how you found out that the sizeof(tmp) is 32, I modified your code to the one below which includes a few printfs and some minor changes:
#include "stdio.h"
#include "stdlib.h"
#include "string.h"
void reverseString(char *p_string)
{
size_t length = strlen(p_string);
size_t r_it = length - 1;
char* tmp = (char*)malloc(length+1);
int last_it = 0;
size_t i=0;
printf("strlen(p_string) = %d\n", strlen(p_string));
printf("Before: strlen(tmp) = %d\n", strlen(tmp));
for (i = 0; i != length; i++) {
tmp[i] = p_string[r_it];
r_it--;
last_it++;
}
tmp[last_it] = '\0';
printf("After: strlen(tmp) = %d\n", strlen(tmp));
strcpy(p_string, tmp);
free(tmp);
}
int main()
{
char str[] = "StackOverflow";
reverseString(str);
printf("%s\n", str);
return 0;
}
First, I have removed all C++ specific code - you can now compile this with gcc. Running this code yields this output:
sizeof(p_string) = 13
Before: strlen(tmp) = 0
After: strlen(tmp) = 13
wolfrevOkcatS
This is to be expected - strlen basically counts bytes until it hits the \0 character and so the first time we print the size using strlen, it returns 0 since we just allocated the memory. As another poster suggested, we have to allocate 1 extra byte to store the \0 in our new string.
Once the reverse is complete, 13 bytes would have been copied over to this memory and the second strlen returns the expected answer.
/* program to accept and print 5 strings using pointers */
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define SIZE 100
int main()
{
char **s;
int i;
s = (char *)malloc(sizeof(char)*5);
for(i=0;i<5;i++)
{
*(s+i) = (char *)malloc(sizeof(char)*SIZE);
}
printf("enter 5 strigs\n");
for(i = 0;i<5;i++)
{
fgets((s+i),SIZE,stdin);
}
//printing the strings
for(i = 0;i<5;i++)
{
puts((s+i));
}
return 0;
}
This program accepts 5 strings from keyboard and prints on screen.It works properly but shows many warnings.Is there any other ways to do same operation(using pointers only).please suggest me.
The biggest issue in your code is that you do not allocate enough memory to s:
s = malloc(sizeof(char*)*5);
// ^
// Add an asterisk here
Otherwise, it's undefined behavior.
Second biggest is that you are not freeing any of the memory that you allocated. This is a memory leak. To do it properly you need a loop that calls free on every element of s, and then on the s itself:
for(i = 0;i<5;i++) {
free(s[i]);
}
free(s);
Finally, you should not be casting the results of malloc.
As a point for style, consider defining a constant for the number of elements in s, in the same way that you defined SIZE instead of using 100 directly. It may also make sense to switch to array-style dereference, rather than doing pointer arithmetic manually, i.e. puts(s[i]); instead of puts((s+i));
int main()
{
char **s;
int i;
s = (char **)malloc(sizeof(char*)*5); //double pointer (char**)
for(i=0;i<5;i++)
{
*(s+i) = (char *)malloc(sizeof(char)*SIZE);
}
printf("enter 5 strigs\n");
for(i = 0;i<5;i++)
{
fgets(*(s+i),SIZE,stdin); // first location *(s+0)
}
//printing the strings
for(i = 0;i<5;i++)
{
puts(*(s+i)); // first location *(s+0)
}
return 0;
}
I am still new with C and I am trying to empty a 2d char array. Here is the declaration:
char arg_array = (char**)calloc(strlen(buf), sizeof (char**));
for(i = 0; i<(strlen(buf)); i++)
{
arg_array[i] = (char*) calloc (strlen(buf), sizeof(char*));
}
Here is where I try to empty it:
void make_empty(char **arg_array)
{
int i;
for(i = 0; i <= BUFSIZ; i++)
{
arg_array[i][0] = '\0';
}
return;
}
Any help is appreciated
So, am I doing it right because this seems to give me segfaults when I try to add data to the array again and then print it?
Empty is just to have it empty - how can I explain more? lol
Try this:
void make_empty(char **arg_array, int rows, int cols)
{
int i,j;
for(i = 0; i <rows; i++)
{
for(j=0; j<cols;j++)
{
arg_array[i][j] = '\0';
}
}
return;
}
Where rows is number of rows and cols number of cols of your array.
P.S. This function clears the whole array as you should always do. As I commented before, putting '\0' as a first char in string does not clear the whole row, it only makes the rest of it ,,invisible'' for functions like printf. Check this link for more information: http://cplusplus.com/reference/clibrary/cstdio/printf/
There is no need to empty it. Often in C, memory allocation is done with malloc which simply returns to you a block of memory which is deemed owned by the caller. When calloc is called, as well as returning you a block of memory, the memory is guaranteed to be initialized to 0. This means for all intents and purposes it is already 'empty'.
Also I'm not quite sure if your code does what you are intending. Let me explain what it does at the moment:
char arg_array = (char**)calloc(strlen(buf), sizeof (char**));
This line is simply wrong. In C, there is no need to cast pointers returned from calloc because they are of type void *, which is implicitly casted to any other pointer type. In this case, you are storing it in a char type which makes no sense. If you do this:
char ** arg_array = calloc(strlen(buf), sizeof (char**));
Then it allocates an array of pointers of strlen(buf) length. So if buf is "hello" then you have now allocated an array which can store 5 pointers.
for(i = 0; i<(strlen(buf)); i++)
{
arg_array[i] = calloc (strlen(buf), sizeof(char*));
}
Again, I have removed the redundant cast. What this does is populates the array allocated earlier. Each index of the array now points to a char string of strlen(buf) * sizeof(char *) length. This is probably not what you want.
Your question is more clear to me now. It appears you want to remove the strings after populating them. You can do it two ways:
Either free each of the pointers and allocate more space later as you did before
Or set the first character of each of the strings to a null character
To free the pointers:
for(i = 0; i<(strlen(buf)); i++)
{
free(arg_array[i]);
}
To set the first character of each string to a null character:
for(i = 0; i<(strlen(buf)); i++)
{
arg_array[i][0] = '\0';
}
That is the same code as what you have originally and should be fine.
As proof, the following code will run without errors:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main(void)
{
char * buf = "hello";
char ** arg_array = calloc(strlen(buf), sizeof (char**));
unsigned int i;
for(i = 0; i < strlen(buf); i++) {
arg_array[i] = calloc(strlen(buf),
sizeof(char *));
}
for(i = 0; i < strlen(buf); i++) {
arg_array[i][0] = '\0';
}
for(i = 0; i < strlen(buf); i++) {
free(arg_array[i]);
}
free(arg_array);
return EXIT_SUCCESS;
}
If your code is segfaulting, the problem is coming from somewhere else. Did you overwrite the arg_array variable? Are you sure BUFSIZE is equal to strlen(buf)?