This program is giving wrong output,, basically i want to remove the character specified and replace it by 'g'...For e.g: All that glitters is not gold if the user entered o then the output should be All that glitters is ngt ggld but the program is deleting all the characters from n onwards
#include <iostream>
using namespace std;
int main()
{
string input(" ALL GLItters are not gold");
char a;
cin>>a;
for(int i=0;i<input.size();i++)
{
if(input.at(i)==a)
{
input.erase(i,i+1);
input.insert(i,"g");
}
}
cout<<"\n";
cout<<input;
}
string& erase (size_t pos = 0, size_t len = npos);
The second parameter ( len ) is the Number of characters to erase.
You have to put 1 not i+1 :
input.erase(i,1);
http://www.cplusplus.com/reference/string/string/erase/
Why not replace it directly? Replace your for loop with this:
for (char& c : input)
{
if (c == a)
c = 'g';
}
Live example here.
Related
When I run the code using the following key, extra characters are outputted...
TERMINAL WINDOW:
$ ./substitution abcdefghjklmnopqrsTUVWXYZI
plaintext: heTUXWVI ii ssTt
ciphertext: heUVYXWJ jj ttUuh|
This is the instructions (cs50 substitution problem)
Design and implement a program, substitution, that encrypts messages using a substitution cipher.
Implement your program in a file called substitution.c in a ~/pset2/substitution directory.
Your program must accept a single command-line argument, the key to use for the substitution. The key itself should be case-insensitive, so whether any character in the key is uppercase or lowercase should not affect the behavior of your program.
If your program is executed without any command-line arguments or with more than one command-line argument, your program should print an error message of your choice (with printf) and return from main a value of 1 (which tends to signify an error) immediately.
If the key is invalid (as by not containing 26 characters, containing any character that is not an alphabetic character, or not containing each letter exactly once), your program should print an error message of your choice (with printf) and return from main a value of 1 immediately.
Your program must output plaintext: (without a newline) and then prompt the user for a string of plaintext (using get_string).
Your program must output ciphertext: (without a newline) followed by the plaintext’s corresponding ciphertext, with each alphabetical character in the plaintext substituted for the corresponding character in the ciphertext; non-alphabetical characters should be outputted unchanged.
Your program must preserve case: capitalized letters must remain capitalized letters; lowercase letters must remain lowercase letters.
After outputting ciphertext, you should print a newline. Your program should then exit by returning 0 from main.
My code:
#include <cs50.h>
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(int argc,string argv[])
{
char alpha[26] = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
string key = argv[1];
int totalchar = 0;
for (char c ='a'; c <= 'z'; c++)
{
for (int i = 0; i < strlen(key); i++)
{
if (tolower(key[i]) == c)
{
totalchar++;
}
}
}
//accept only singular 26 key
if (argc == 2 && totalchar == 26)
{
string plaint = get_string("plaintext: ");
int textlength =strlen(plaint);
char subchar[textlength];
for (int i= 0; i< textlength; i++)
{
for (int j =0; j<26; j++)
{
// substitute
if (tolower(plaint[i]) == alpha[j])
{
subchar[i] = tolower(key[j]);
// keep plaintext's case
if (plaint[i] >= 'A' && plaint[i] <= 'Z')
{
subchar[i] = (toupper(key[j]));
}
}
// if isn't char
if (!(isalpha(plaint[i])))
{
subchar[i] = plaint[i];
}
}
}
printf("ciphertext: %s\n", subchar);
return 0;
}
else
{
printf("invalid input\n");
return 1;
}
}
strcmp compares two strings. plaint[i] and alpha[j] are chars. The can be compared with "regular" comparison operators, like ==.
I am trying to write a code that prints initials of a given name? I have been getting error whenever I use '/0' - to initialize the ith character of a given string. I am using it to identify initials of the 2nd word? Any suggestions to detect the 2nd word and print the initial of the 2nd word? Additionally, I am trying to write this code so that it ignores any additional spaces:
#include <stdio.h>
#include <cs50.h>
#include <string.h>
#include <ctype.h>
int main(void)
{
printf ("Enter you Name:");//print name
string s = get_string(); //get input from user
printf("%c", toupper(s[0])); // use placeholder to call the string
for (int i=0, n = strlen(s); i < n; i++)
{
if (s[i] == '/0')
{
printf("%c", toupper(s[i+1]));
}
}
printf ("\n");
}
I'm not sure what your get_string() function does but this does what you're asking for.
int main(void)
{
printf ("Enter you Name:");//print name
char input[100];
cin.getline(input, sizeof(input));
string s(input);
//get input from user
printf("%c", toupper(s[0])); // use placeholder to call the string
for (int i=0, n = s.length(); i < n; i++)
{
if (isspace(s.at(i)))
{
printf("%c", toupper(s[i+1]));
}
}
printf ("\n");
}
Try to use ' ' instead of '/0'.
if (s[i] == ' ')
or if you prefer ASCII codes (Space is 0x20 or 32 as decimal):
if (s[i] == 0x20)
your code will work if there is only 1 space between names. To avoid this, the condition should also check the next char. If it's not a space then probably it's a 'letter':
if (s[i] == 0x20 && s[i+1] != 0x20) {...
Note that the for cycle should stop at i < n - 1, otherwise the loop will fail if there are trailing spaces...
My doubts are as follows :
1 : how to send 'str' from function 'fun' , So that i can display it in main function.
2 : And is the return type correct in the code ?
2 : the current code is displaying some different output.
char * fun(int *arr)
{
char *str[5];
int i;
for(i=0;i<5;i++)
{
char c[sizeof(int)] ;
sprintf(c,"%d",arr[i]);
str[i] = malloc(sizeof(c));
strcpy(str[i],c);
}
return str;
}
int main()
{
int arr[] = {2,1,3,4,5},i;
char *str = fun(arr);
for(i=0;i<5;i++)
{
printf("%c",str[i]);
}
return 0;
}
how to send 'str' from function 'fun' , So that i can display it in main function.
This is the way:
char* str = malloc( size );
if( str == NULL ) {
fprintf( stderr,"Failed to malloc\n");
}
/* Do stuff with str, use str[index],
* remember to free it in main*/
free(str);
And is the return type correct in the code ?
No, Probably char** is the one you need to return.
the current code is displaying some different output.
Consider explaining what/why do you want to do ? The way you have written, seems completely messed up way to me. You're passing array of integer but not its length. How is the fun() supposed to know length of array? Another problem is array of pointers in fun().
You can't write a int to a char (See the both size). So I used char array instead.
However, I'm not sure if this is what you want to do (might be a quick and dirty way of doing it):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char**
fun(int *arr, int size)
{
char **str = malloc( sizeof(char*)*size );
if( str == NULL ) {
fprintf( stderr, "Failed malloc\n");
}
int i;
for(i=0;i<5;i++) {
str[i] = malloc(sizeof(int));
if( str == NULL ) {
fprintf( stderr, "Failed malloc\n");
}
sprintf(str[i],"%d",arr[i]);
}
return str;
}
int
main()
{
int arr[] = {2,1,3,4,5},i;
char **str = fun(arr, 5);
for(i=0;i<5;i++) {
printf("%s\n",str[i]);
free(str[i]);
}
free(str);
return 0;
}
I made these changes to your code to get it working:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char **fun(int *arr)
{
char **str = malloc(sizeof(char *) * 5);
int i;
for(i = 0; i < 5; i++) {
if ((arr[i] >= 0) && (arr[i] <= 9)) {
char c[2] ;
sprintf(c, "%d", arr[i]);
str[i] = (char *) malloc(strlen(c) + 1);
strcpy(str[i],c);
}
}
return str;
}
int main()
{
int arr[] = {2, 1, 3, 4, 5}, i;
char **str = fun(arr);
for(i = 0; i < 5; i++) {
printf("%s", str[i]);
free(str[i]);
}
printf("\n");
free(str);
return 0;
}
Output
21345
I added a check to make sure that arr[i] is a single digit number. Also, returning a pointer to a stack variable will result in undefined behavior, so I changed the code to allocate an array of strings. I don't check the return value of the malloc calls, which means this program could crash due to a NULL pointer reference.
This solution differs from the others in that it attempts to answer your question based on the intended use.
how to send 'str' from function 'fun' , So that i can display it in main function.
First, you need to define a function that returns a pointer to array.
char (*fun(int arr[]))[]
Allocating variable length strings doesn't buy you anything. The longest string you'll need for 64bit unsigned int is 20 digits. All you need is to allocate an array of 5 elements of 2 characters long each. You may adjust the length to suit your need. This sample assumes 1 digit and 1 null character. Note the allocation is done only once. You may choose to use the length of 21 (20 digits and 1 null).
For readability on which values here are related to the number of digits including the terminator, I'll define a macro that you can modify to suit your needs.
#define NUM_OF_DIGITS 3
You can then use this macro in the whole code.
char (*str)[NUM_OF_DIGITS] = malloc(5 * NUM_OF_DIGITS);
Finally the receiving variable in main() can be declared and assigned the returned array.
char (*str)[NUM_OF_DIGITS] = fun(arr);
Your complete code should look like this:
Code
char (*fun(int arr[]))[]
{
char (*str)[NUM_OF_DIGITS] = malloc(5 * NUM_OF_DIGITS);
int i;
for(i=0;i<5;i++)
{
snprintf(str[i],NUM_OF_DIGITS,"%d",arr[i]); //control and limit to single digit + null
}
return str;
}
int main()
{
int arr[] = {24,1,33,4,5},i;
char (*str)[NUM_OF_DIGITS] = fun(arr);
for(i=0;i<5;i++)
{
printf("%s",str[i]);
}
free(str);
return 0;
}
Output
2413345
With this method you only need to free the allocated memory once.
I am working on the Vigenere exercise from Harvard's CS50 (in case you noticed I'm using string and not str).
My program gives me a Floating Point Exception error when I use "a" in the keyword.
It actually gives me that error
when I use "a" by itself, and
when I use "a" within a bigger word it just gives me wrong output.
For any other kind of keyword, the program works perfectly fine.
I've run a million tests. Why is it doing this? I can't see where I'm dividing or % by 0. The length of the keyword is always at least 1. It is probably going to be some super simple mistake, but I've been at this for about 10 hours and I can barely remember my name.
#include <stdio.h>
#include <cs50.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
int main (int argc, string argv[])
{
//Error message if argc is not 2 and argv[1] is not alphabetical
if (argc != 2)
{
printf("Insert './vigenere' followed by an all alphabetical key\n");
return 1;
}
else if (argv[1])
{
for (int i = 0, n = strlen(argv[1]); i < n; i++)
{
if (isalpha((argv[1])[i]) == false)
{
printf("Insert './vigenere' followed by an all alphabetical key\n");
return 1;
}
}
//Store keyword in variable
string keyword = argv[1];
//Convert all capital chars in keyword to lowercase values, then converts them to alphabetical corresponding number
for (int i = 0, n = strlen(keyword); i < n; i++)
{
if (isupper(keyword[i])) {
keyword[i] += 32;
}
keyword[i] -= 97;
}
//Ask for users message
string message = GetString();
int counter = 0;
int keywordLength = strlen(keyword);
//Iterate through each of the message's chars
for (int i = 0, n = strlen(message); i < n; i++)
{
//Check if ith char is a letter
if (isalpha(message[i])) {
int index = counter % keywordLength;
if (isupper(message[i])) {
char letter = (((message[i] - 65) + (keyword[index])) % 26) + 65;
printf("%c", letter);
counter++;
} else if (islower(message[i])) {
char letter = (((message[i] - 97) + (keyword[index])) % 26) + 97;
printf("%c", letter);
counter++;
}
} else {
//Prints non alphabetic characters
printf("%c", message[i]);
}
}
printf("\n");
return 0;
}
}
This behavior is caused by the line keyword[i] -= 97;, there you make every 'a' in the key stream a zero. Later you use strlen() on the transformed key. So when the key starts with an 'a', keywordLength therefor is set to zero, and the modulo keywordLength operation get into a division by zero. You can fix this by calculating the keyword length before the key transformation.
I've come across this
#define DsHook(a,b,c) if (!c##_) { INT_PTR* p=b+*(INT_PTR**)a; VirtualProtect(&c##_,4,PAGE_EXECUTE_READWRITE,&no); *(INT_PTR*)&c##_=*p; VirtualProtect(p,4,PAGE_EXECUTE_READWRITE,&no); *p=(INT_PTR)c; }
and everything is clear except the "c##_" word, what does that mean?
It means to "glue" together, so c and _ get "glued together" to form c_. This glueing happens after argument replacement in the macro. See my example:
#define glue(a,b) a##_##b
const char *hello_world = "Hello, World!";
int main(int arg, char *argv[]) {
printf("%s\n", glue(hello,world)); // prints Hello, World!
return 0;
}
It is called a token-pasting operator. Example:
// preprocessor_token_pasting.cpp
#include <stdio.h>
#define paster( n ) printf( "token" #n " = %d", token##n )
int token9 = 9;
int main()
{
paster(9);
}
Output
token9 = 9
That's concatenation that appends an underscore to the name passed as c. So when you use
DsHook(a,b,Something)
that part turns into
if (!Something_)
After the preprocessor, your macro will be expanded as:
if (!c_) { INT_PTR* p=b+*(INT_PTR**)a; VirtualProtect(&c_,4,PAGE_EXECUTE_READWRITE,&no); *(INT_PTR*)&c_=*p; VirtualProtect(p,4,PAGE_EXECUTE_READWRITE,&no); *p=(INT_PTR)c; }
The ## directive concatenates the value of c which you pass as a macro parameter to _
Simple one:
#define Check(a) if(c##x == 0) { }
At call site:
int varx; // Note the x
Check(var);
Would expand as:
if(varx == 0) { }
It is called Token Concatenation and it is used to concatenate tokens during the preprocessing
For example the following code will print out the values of the values of c, c_, c_spam:
#include<stdio.h>
#define DsHook(a,b,c) if (!c##_) \
{printf("c=%d c_ = %d and c_spam = %d\n",\
c, c##_,c##_spam);}
int main(){
int a,b,c=3;
int c_ = 0, c_spam = 4;
DsHook(a,b,c);
return 0;
}
Output:
c=3 c_ = 0 and c_spam = 4