passing char* args[] via voidpointer - c

Her is my simplified code:
void main(){
void* ptr;
char* args[3];
args[0]="Arg1";
args[1]="Arg2";
args[2]="Arg3";
ptr = &args;
myMethod(ptr);
}
static void myMethod(void* args){
}
how can I convert void* args to an char*[]? in myMethod(void*)?

You want a pointer to pointer to char:
#include <stdio.h>
static void myMethod(void *args)
{
char **ptr = args;
printf("%s\n", ptr[1]);
}
int main(void)
{
void *ptr;
char *args[3];
args[0]="Arg1";
args[1]="Arg2";
args[2]="Arg3";
ptr = args; /* You don't need the & */
myMethod(ptr);
return 0;
}
As pointed out by #Eregrith, pass the number of cells to the function in order to prevent an out of bounds access:
#include <stdio.h>
static void myMethod(void *args, size_t elems)
{
char **ptr = args;
for (size_t i = 0; i < elems; i++)
printf("%s\n", ptr[i]);
}
int main(void)
{
void *ptr;
char *args[3];
args[0]="Arg1";
args[1]="Arg2";
args[2]="Arg3";
ptr = args;
myMethod(ptr, sizeof(args) / sizeof(args[0]));
return 0;
}

Related

How to convert array of char into array of int?

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "stats.h"
/* Size of the Data Set */
#define SIZE (40)
void print_array (unsigned char *p, int l) {
int i;
for (i=0;i<l;i++) {
printf("%d\t",*p);
p++;
}
}
void print_array_int (int *p, int l) {
int i;
for (i=0;i<l;i++) {
printf("%d\t",*p);
p++;
}
}
void typecasting(unsigned char test[SIZE], int array[SIZE]) {
int i=0;
unsigned char *token = strtok(test,",");
while (token) {
if(i<SIZE) {
array[i++] = atoi(token);
}
token = strtok(NULL,",");
}
}
void main() {
int array[SIZE] = {};
unsigned char test[SIZE] = {34,201,190,154,8,194,2,6,114,88,45,76,123,87,25,23,200,122,150,90,92,87,177,244,201,6,12,60,8,2,5,67,7,87,250,230,99,3,100,90};
/* Other Variable Declarations Go Here */
/* Statistics and Printing Functions Go Here */
print_array(test, SIZE);
typecasting(test,array);
print_array_int(array,SIZE);
}
What I want in this code is to convert the array of char into an array of int.
Previously I tried doing this by using pointers but didn't work and it showed stack smashing error. I want to convert this array of char into array of int to perform some mathematical operations.
You are trying too hard. Here's how typecasting should look
void typecasting(unsigned char test[SIZE], int array[SIZE]) {
for (int i = 0; i < SIZE; ++i)
array[i] = test[i];
}
Your code might be suitable if you were converting from a C string, i.e. if your original test array was
char test[] = "34,201,190,154,8,194,2,6,114,88,45,76,123,87,25,23,...";
So I guess you could say you're misunderstanding the nature of char (and unsigned char) in C++. They can represent character data as in char greeting[] = "hello"; or they can represent small integers as in char test[] = {1,2,3};.

how can I iterate through an array of generic type (void**) in with function pointer in C?

I want to perform a simple task working with function pointers in C.
The task is to get an array (from any type, i.e: int / char*), and sum /concatenate each 2 elements in the array.
for the char* type, it works fine, but for the int type, the loop seems to jump each 2 elements in the array (and thus overflow the array):
#define N1 4
#define N2 4
typedef void*(*Fn_Sum)(void*, void*);
typedef void(Fn_Prt)(void*);
int sum_num(int a, int b){
return a + b;
}
char* sum_char(char* a, char* b){
char *result = malloc(strlen(a) + strlen(b) + 1);
if (!result) {
printf("ERROR: malloc failed !\n");
return NULL;
}
strcpy(result, a);
strcat(result, b);
return result;
}
void print_num(int a){
printf("%d", a);
}
void print_string(char* a){
int i = 0;
while (a[i] != '\0') {
printf("%c", a[i]);
i++;
}
}
void PrintSums(void** P, int n, Fn_Sum fsum, Fn_Prt fprt){
for(int i = 0; i < n - 1; i++){
(fprt)(fsum(P[i], P[i+1]));
printf(", ");
}
printf("\n");
}
int main() {
int V[N1] = {1,2,3,4};
char* S[N2] = {"a", "d", "c", "d"};
PrintSums(V, N1, sum_num, print_num);
PrintSums(S, N2, sum_char, print_string);
return 0;
}
expected output is :
3, 5, 7,
ab, bc, cd,
actual outputs:
4, 725939, 4925336,
ad, dc, cd,
Create an abstract interface for iterator over the elements. A draft of such interface could look like this:
struct iterator {
...
};
// ptr - a pointer to beginning of the array
// size - size of one element in the array
void it_init(iterator *t, void *ptr, size_t size);
bool it_eq(iterator *t, iterator *o); // compare iteratores
void it_add(iterator *t, size_t n);
void it_inc(iterator *t);
// return a pointer to the element
void *it_get(iterator *t);
Remember to always pass to user callbacks a context variables. Otherwise users will have to use global variables, which make code messy. Create an abstract interface with destructors and constructors of your summing object. Handle errors properly:
// is passed a pointer to user context
// returns 0 on success
typedef int (*Fn_Sum)(void*, void*);
// is passed a pointer to user context
// returns 0 on success
typedef int (Fn_Prt)(void*);
// returns 0 on success
int PrintSums(iterator it, size_t n, Fn_Sum fsum, Fn_Prt fprt, void *sumctx);
After that, implement objects that expose the interface that you want:
struct num { .. };
void num_sum(struct num *t, int el);
void num_print(struct num *t, int el);
// expose interface to PrintSums
// that just calls internal api
int num_PrintSums_Fn_Sum(void *ctx, void *el0) {
struct num *t = ctx;
int *el = el0;
num_sum(t, *el);
return 0;
}
int num_PrintSums_Fn_Prt(void *ctx) {
struct num *t = ctx;
num_print(t);
return 0;
}
An example whole program looks like this:
#include <stddef.h>
#include <stdbool.h>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
// iterator
typedef struct iterator {
void *ptr;
size_t size;
} iterator;
void it_init(iterator *t, void *ptr, size_t size) {
*t = (iterator){ ptr, size, };
}
// eq is from test(1) shell command. "eq" means "equal"
bool it_eq(iterator *t, iterator *o) {
return t->ptr == o->ptr;
}
void it_add(iterator *t, size_t n) {
t->ptr = (char*)t->ptr + t->size * n;
}
// increment the iterator
void it_inc(iterator *t) {
it_add(t, 1);
}
// return a pointer to the element
void *it_get(iterator *t) {
return t->ptr;
}
// interface
typedef int (*Fn_Sum)(void*, void*);
typedef int (Fn_Prt)(void*);
int PrintSums(iterator it, size_t n, Fn_Sum fsum, Fn_Prt fprt, void *sumctx){
iterator end = it;
it_add(&end, n);
for(; !it_eq(&it, &end); it_inc(&it)) {
int err = fsum(sumctx, it_get(&it));
if (err) return err;
err = fprt(sumctx);
if (err) return err;
printf(", ");
}
printf("\n");
return 0;
}
// num object
struct num {
int sum;
};
void num_init(struct num *t) {
t->sum = 0;
}
void num_sum(struct num *t, int el){
t->sum += el;
}
void num_print(struct num *t){
printf("%d", t->sum);
}
void num_free(struct num *T) {
// nothing, just exists for uniform API
}
// accessors for PrintSums
int num_PrintSums_Fn_Sum(void *ctx, void *el0) {
struct num *t = ctx;
int *el = el0;
num_sum(t, *el);
return 0;
}
int num_PrintSums_Fn_Prt(void *ctx) {
struct num *t = ctx;
num_print(t);
return 0;
}
// string object
struct str {
char *str;
};
void str_init(struct str *t) {
t->str = NULL;
}
int str_sum(struct str *t, const char *str) {
const size_t str_len = t->str == NULL ? 0 : strlen(t->str);
void *p = realloc(t->str, str_len + strlen(str) + 1);
if (p == NULL) {
free(t->str);
t->str = NULL;
return -1;
}
t->str = p;
memcpy(t->str + str_len, str, strlen(str) + 1);
return 0;
}
void str_print(struct str *t) {
if (t->str == NULL) {
printf("(nul)");
} else {
printf("%s", t->str);
}
}
void str_free(struct str *t) {
free(t->str);
}
// interface for PrintSums
int str_PrintSums_Fn_Sum(void *ctx, void *el0) {
struct str *t = ctx;
const char **el = el0;
str_sum(t, *el);
return 0;
}
int str_PrintSums_Fn_Prt(void *ctx) {
struct str *t = ctx;
str_print(t);
return 0;
}
// and finally main
int main() {
int err = 0;
int V[] = {1,2,3,4};
iterator numit;
it_init(&numit, V, sizeof(*V));
struct num numsum; // the object that will hold the sum
num_init(&numsum);
err = PrintSums(numit, sizeof(V)/sizeof(*V), num_PrintSums_Fn_Sum, num_PrintSums_Fn_Prt, &numsum);
if (err) abort();
num_free(&numsum);
char *S[] = {"a", "d", "c", "d"};
iterator strit;
it_init(&strit, S, sizeof(*S));
struct str strsum; // the object that will hold the sum of strings
str_init(&strsum);
err = PrintSums(strit, sizeof(S)/sizeof(*S), str_PrintSums_Fn_Sum, str_PrintSums_Fn_Prt, &strsum);
if (err) abort();
str_free(&strsum); // YES! Remember to pick out the trash
}
and outputs on godbolt:
1, 3, 6, 10,
a, ad, adc, adcd,
The pointers to a constructor and destructor of "sum objects" could be passed to PrintSums too. That said one could start thinking about creating a virtual table for all these pointers (ie. one struct with function pointers that are needed for PrintSums...).

How to pass 2d array of string to the function and print value of it?

Why it is not working... It should be working, right? gcc have problem with this line, but why?
render_history(history, 2);
Sorry for bothering. I am just a beginner.
#include <stdio.h>
void render_history(char** history, const int entry);
int main()
{
char* history[3][4];
history[0][0] = "1234";
history[1][0] = "5678";
history[2][0] = "9012";
render_history(history, 2); //??
return 0;
}
void render_history(char** history, const int entry)
{
// print "9012"
}
gcc have problem with this line, but why?
Because the type is wrong. char* history[3][4]; can't be passed as char**. They are incompatible types.
Try something like:
#include <stdio.h>
void render_history(char* (*history)[4] , const int entry)
{
printf("%s\n", history[entry][0]);
}
int main()
{
char* history[3][4];
history[0][0] = "1234";
history[1][0] = "5678";
history[2][0] = "9012";
render_history(history, 2);
return 0;
}
As mentioned above double pointer not equal to 2D array.
You can also use pointer to pointer of char. char **history. And with this you have several option:
1) Use compound literals
#include <stdio.h>
void render_history(const char **history, const int entry)
{
printf("%s\n", history[entry]);
}
int main(void)
{
const char **history = (const char *[]) { "1234", "5678", "9012", NULL};
render_history(history, 2);
return 0;
}
If you need change your data later
2) Use dynamic memory allocation with malloc
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void render_history(char **history, const int entry)
{
printf("%s\n", history[entry]);
}
int main(void)
{
char **history = malloc(3 * sizeof(char *));
for (int i = 0; i < 3; ++i)
{
history[i] = malloc(4 * sizeof(char));
}
strcpy(history[0], "1234");
strcpy(history[1], "5678");
strcpy(history[2], "9012");
history[3] = NULL;
render_history(history, 2);
return 0;
}
If you use 2nd option dont forget free memory after use.

C pointers weird behaviour

I am having trouble comprehending why this works:
int main() {
int test = 4;
int *bar = &test;
int **out = &bar;
printf("%d\n", **out);
return 0;
}
but this doesn't:
void foo(int *src, int **out) {
out = &src;
}
int main() {
int test = 4;
int *bar = &test;
int **out;
foo(bar, out);
printf("%d\n", **out);
return 0;
}
The second snippet throws "Segmentation fault". To me it seems they do the same thing. Can someone explain please?
Edit: (updated code based on answers):
void foo(int *src, int **out) {
out = &src;
}
int main() {
int test = 4;
int *bar = &test;
int *out;
foo(bar, &out);
printf("%d\n", *out);
return 0;
}
Then why does this not work?
Solved: (I had to think through what I really wanted to do), this is the result:
void foo(int *src, int **out) {
*out = src;
}
int main() {
int test = 4;
int *bar = &test;
int *out;
foo(bar, &out);
printf("%d\n", *out);
return 0;
}
In the second, the variable out in main is not affected by the assignment made inside of foo.
In your edit, you need to have foo assign to what out in it points to:
*out = src;
Things are passed into functions by value. Changing the value in the function changes nothing in the outside world.

Wrong output in my stringreverse program. what is fault in my code?

#include<stdio.h>
#include<conio.h>
#include<string.h>
char* strreverse(char*);
int main()
{
char *rev_string;
char *name="computer";
clrscr();
rev_string=strreverse(name);
printf("%s", rev_string);
getch();
return 0;
}
char* strreverse(char *name)
{
int length=strlen(name);
char *ptr;
char *rstr;
for(ptr=name+(length-1);ptr>=name;ptr--)
{
*rstr=*ptr;
printf("%c",rstr);
rstr++;
}
*(rstr)=NULL;
return rstr;
}
the above is my code. i tried to write a program for string reverse without using arrays. But i am not getting the output retupmoc. what is wrong in my code? how to insert null char in char*?
#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <stdlib.h>
char* strreverse(const char*);
int main(){
char *rev_string;
char *name="computer";
clrscr();
rev_string=strreverse(name);
printf("%s\n", rev_string);
free(rev_string);
getch();
return 0;
}
char* strreverse(const char *name){
int length=strlen(name);
const char *ptr;
char *ret, *rstr = malloc(length + 1);
if(ret=rstr){
for(ptr=name+length;ptr != name;){
*rstr++ = *--ptr;
}
*rstr = '\0';
}
return ret;
}
You did not allocate memory to hold your reversed string. Try
char *rstr = calloc(1, length+1);
Also it should be
printf("%c", *rstr); // dereference
*(rstr)= '\0'; // instead of NULL
Here you find sweet and short solution for string reverse:
#include<stdio.h>
#include<string.h>
int strreverse(char* , char*);
int main()
{
char rev_string[10] = {0};
char name[10]="computer";
strreverse(name, rev_string);
printf("%s\n", rev_string);
return 0;
}
int strreverse(char *name, char *rStr)
{
int i = 0;
int length = strlen(name);
while(i < length)
{
rStr[i] = name[length-i-1];
i++;
}
return 0;
}
Try to run and have fun.

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