I am trying to create a hexadecimal to base64 converter. I don't know if I am going in the right direction converting binary as I am attempting to do now or if there is a more direct way of converting. Any suggestions on the math of converting or how to code hex to base64 would be very helpful.
On the contrary, I have been receiving the error(The new one since updated code):
hexto64.c: In function ‘main’:
hexto64.c:21:17: error: lvalue required as left operand of assignment
ReVerse(input) = RevHex;
^
If anybody could help explain what this error means and how to fix
would help greatly! Thank you in advance.
EDIT: So thanks to the few people in the comments, I now understand the error.
Here is my code(Updated):
#include <stdio.h>
#include <math.h>
#include <string.h>
char ReVerse(const char *str)
{
if (*str != '\0')
ReVerse((str + 1));
printf("%c", *str);
}
int main()
{
char RevHex;
char input[4096] = {0};
printf("Enter Hexadecimal: ");
scanf("%s", input);
RevHex = ReVerse(input);
printf("\n");
return 0;
}
Last edit: I have found the error in my code. Thank you guys for all the feedback!
You try to use the return value of the function ReVerse() but this function return void. It's "nothing" so you can't assign "nothing" to something. Here you try to put your array RevHex to "nothing". This don't make sense.
If your function is just reversing the entered Hexadecimal string then you can do it like this:
#include <stdio.h>
#include <math.h>
#include <string.h>
void ReVerse(const char *str) {
if (*str != '\0')
ReVerse((str + 1));
printf("%c", *str); }
int main() {
char input[10] = {0};
printf("Enter Hexadecimal: ");
scanf("%s", input);
ReVerse(input);
printf("\n");
return 0; }
Here I do not think that this char RevHex[4096] = {0}; is useful. Because you want to reverse the input string. If you implement it like this then this program will work.
Talking about your program, you are assigning a string value to a void function, that is why it is giving the error.
I don't know how to "fix" this but trying to assign RevHex to a function will throw an error.
Given that ReVerse doesn't return anything, I am not sure what RevHex is for...
The error is this line: ReVerse(input) = RevHex;
You can't assign something to a call of void function, you should pass it to the function as parameter.
Related
I'm trying to acquaint myself with the atoi function, so I wrote a basic programme using it, but I'm having some problems:
#include <stdio.h>
#include <cs50.h>
#include <stdlib.h>
int main(void)
{
string s = get_string("String:");
int i = get_int("Integer:");
int a = atoi(s[1]);
int j = i + a;
printf("%i\n", j);
}
When I try to compile it, I get the error message "incompatible integer to pointer conversion passing 'char' to parameter of type 'const char *'; take the address with & [-Werror,-Wint-conversion]". This seems to suggest that it wants something to do with a char, but from what I've read, I was under the impression that atoi was used with strings. If someone could explain where I'm going wrong, I'd be very thankful
You're passing a char(assuming string is a typedef for char*) by indexing the string and it wants you to pass a char*. So just pass the full string:
#include <stdio.h>
#include <cs50.h>
#include <stdlib.h>
int main(void)
{
string s = get_string("String:");
int i = get_int("Integer:");
int a = atoi(s);
int j = i + a;
printf("%i\n", j);
}
here is the syntax for atoi()
int atoi(const char *nptr);
Notice the parameter is a pointer, not a single character.
However, the posted code has:
int a = atoi(s[1]);
which is a single character, not a pointer
suggest:
int a = atoi( s );
Also, the function: atoi() has no facility to let the program know when an error occurred. Suggest using the strtol() function which does have a facility to indicate when an error occurred
I've looked around everywhere and tried pretty much everything suggested and can't get anything to work.
this is my code:
#include <stdio.h>
#include <string.h>
#include <math.h>
int main(){
float a;
char *nums[3];
char str[5];
printf("Please enter a,b,c:");
scanf("%s",str);
int i=0;
char *p;
p = strtok (str,",");
while (p != NULL)
{
nums[i++] = p;
p = strtok (NULL, ",");
}
a=atof(nums[0]);
printf("%s\n",nums[0]);
printf("%f\n",a);
return 0;
}
the math.h is for something later on after I figure this out. So if I entered "1,2,3" into this program, my print statements would show me "1" and then "0.000", this is obviously just there for me to test things out but why the does my value just disappear after trying to convert to a float? I need that value of 1 to do math with later in my program but I can't get that char pointer value no matter what I try, I can only seem to print it, but it screws up as soon as I try to convert it into a type I can use.
Two issues:
You nums and str arrays are too short. nums should have a size of at least 3, and str should be at least 6 ("1,2,3" plus null byte), probably more for larger numbers.
So change those to:
char *nums[3];
char str[20];
Second, you don't #include <stdlib.h>, which contains the declaration of atof. Without a declaration, it is assumed to return an int.
Fix the array sizes, and #include <stdlib.h>, and it should work.
I finished my programming classes in C, and thought I would write some code down. BOOM! I run into so many problems. I guess the C language is so complicated, even a book can't explain how it works entirely.
This is my problem(I am trying to display something using a pointer)
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
void main()
{
char *a;
system("cls");
*a = {"hello"};
printf("%s",a);
getch();
}
*a = {"hello"};
This dereferenced the pointer and assigned something to that memory location. The reason it crashed is because the pointer was uninitialised, ie it did not point at anything (actually it did point at something, but that something was an undefined memory location).
If you had done the following, your program would have worked:
a = "hello";
The type of a is char*. The type of "hello" is also char*.
You don't give value to a pointer to char this way:
*a = {"hello"};
You have to use:
a="hello";
I don't understand very well what you are trying to do. If you only want to print "hello" in your screen, why do you use a pointer to char? What is the getch() for? You use that function this way: http://linux.die.net/man/3/getch Do you intend to read a character?
I would only do:
#include <stdio.h>
main()
{
printf("hello");
}
What are you exactly trying to do?
This is a good reference guide: http://www.gnu.org/software/gnu-c-manual/gnu-c-manual.html
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
void main()
{
char *a;// a is pointer which address is store in stack and when u initialize with any string then string is store in code section which cant be change
clrscr();
a = "hello";// this is the way to store the string but if when u assign while declaring as char *a= hello this will accept Remember hello is store in code where as address of pointer a is store in stack section
printf("%s", a);
getch();
}
~
just providing another insight for you..
just now I tried this in Dev-C++ 5.6.3 and it works..
if you assign the value directly when you are declaring it, it works:
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
int main()
{
system("cls");
char *a = {"hello"};
printf("%s",a);
getch();
return 0;
}
and one more thing, clrscr is not a standard c function (it didn't work in my test), so how about using cls in stdlib like I did.. hope it's useful..
I'm new to the world of C programming, and I as trying to code a primitive, terminal-based version of the "Hangman" game.
One of the steps doing this (or at least the way I am working on), is to create a second char array (next to the original char array that stores the word one needs to guess), filled with "*" for every Char of the original array, and display it. Although the rest of the programming part is not there yet (since I am not finished with it yet), I doubt it is relevant for now (however I allready know how to proceed, that is if I weren't bothered by some error-messages)....
Here's the code I have so far:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
void hiddenArray(char array[]);
char *secretWord;
char *arrayCover;
char letter;
int main(int argc, char *argv[]){
secretWord = "Test";
printf("Welcome to the Hangman!\n\n");
hiddenArray(secretWord);
printf("What is the hidden Word?: %c\n", *arrayCover);
printf("Guess a letter\n");
scanf("%c", &letter);
}
void hiddenArray(char array[]){
int size,i;
size = sizeof(*array);
for (i=0;i<size-1;i++){
*(arrayCover+i) = "*";
}
}
Now I have two issues... the first one:
I don't understand the error message I am getting after compilation:
pendu.c:41:19: warning: incompatible pointer to integer conversion assigning to 'char' from 'char [2]' [-Wint-conversion]
*(arrayCover+i) = "*";
^ ~~~
1 warning generated.
And my second question: the second Array created, filled with "*" is not being displayed, what did I do wrong?
I'd love for some help, cheers!
Your program have some errors to be corrected .
1) Your *arraycover is pointing to some unknown value, You have not initialized it.
2) sizeof(*array) should be sizeof(array)
3) *(arrayCover+i) = "*" should be *(arrayCover+i) = '*';
I suggest you not to create too many global variables when you dont really need them
Create char secretWord[100] = "Test" instead of char *secretWord
Try this
size = sizeof(arrayCover)/sizeof(char);
for (i=0;i<size-1;i++){
*(array+i) = '*';
}
Despite of all these, I think you need to allocate memory for arrayCover
arrayCover = malloc(sizeof(char) * number_of_elements);
"*" is a string. Use '*' instead to get the char.
"*" is a string which also contains \0 also, resulting to 2 characters. Instead use '*' which is just a character.
The function hiddenArray has the formal argument array which need to be used locally instead of arrayCover as below,
void hiddenArray(char array[]){
int size,i;
size = sizeof(array)/sizeof(array[0]);
for (i=0;i<size-1;i++){
*(array+i) = '*'; /* Or array[i] = '*' */
}
}
char *test = "hello";
test = change_test("world");
printf("%s",test);
char* change_test(char *n){
printf("change: %s",n);
return n;
}
im trying to pass a 'string' back to a char pointer using a function but get the following error:
assignment makes pointer from integer without a cast
what am i doing wrong?
A function used without forward declaration will be considered having signature int (...). You should either forward-declare it:
char* change_test(char*);
...
char* test = "hello";
// etc.
or just move the definition change_test before where you call it.
printf() prints the text to the console but does not change n. Use this code instead:
char *change_test(char *n) {
char *result = new char[256];
sprintf(result, "change: %s", n);
return result;
}
// Do not forget to call delete[] on the value returned from change_test
Also add the declaration of change_test() before calling it:
char *change_test(char *n);
You're converting an integer to a pointer somewhere. Your code is incomplete, but at a guess I'd say it's that you're not defining change_test() before you use it, so the C compiler guesses at its type (and assumes it returns an integer.) Declare change_test() before calling it, like so:
char *change_test(char *n);
thanks a bunch guys! didnt think i would have this problem solved by lunch. here is the final test class
/* standard libraries */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* change_test(char*);
int main(){
char *test = "hello";
test = change_test("world");
printf("%s",test);
return (EXIT_SUCCESS);
}
char* change_test(char *n){
return n;
}