I want to make an employee record that inputs these members:
Int I'd
Char name
Car gender
Char phone number
Char address
Char salary
ID must be unique and can take only digits
Name must be alphabetic and can take only four spaces. Otherwise display a message
Gender can only F And M
Phone number can only digits and '-'. This short dash
Salary also can take digits
Please, help me as soon as possible.
You can set conditions based on the type of value.
For example:
final int ids[25];
int position;
//in the getid function
public void getid()
{
int i=0;
int id[position]=scanf("%d",&id);
for(i=20;i<=0;i--){
if(id[position]==ids[i])
{
//then show the message
printf("Duplicate id");
}
}
else{
printf("Nice!!");
}
}
Related
For one of my requirements I need the JSON GENERATE function in COBOL 6. My problem is, that it returns UTF-8, but I need the data in EBCDIC (CCSID 1140). Is there a way to convert this? Every solution I found uses national data types, but I have to use the NODBCS compiler option, so those don't work.
I do apologize for not first asking a question (but I am too new to StackOverflow to allow that.) The question would be "do you have C++ and can you link C++ with your COBOL?" I just tried this program:
#include <iconv.h>
class myConv
{
public:
static myConv globalConv;
size_t conv(char ** restrict f, unsigned int * restrict flen,
char ** restrict t, unsigned int * restrict tlen)
{
if (ok_)
{
return iconv(cd_, f, flen, t, tlen);
}
else
{
return (size_t)-1;
}
}
private:
myConv()
{
cd_ = iconv_open("1047", // EBCDID
"1208"); // UTF-8
ok_ = (cd_ != (iconv_t)-1);
// possibly indicate what the error is
}
~myConv()
{
if (ok_)
{
if (iconv_close(cd_) != 0)
{
// possibly indicate what the error is
}
}
}
bool ok_;
iconv_t cd_;
};
myConv myConv::globalConv;
extern "C" bool CNV(char * f, unsigned int flen,
char * t, unsigned int tlen)
{
return myConv::globalConv.conv(&f, &flen,
&t, &tlen) != (size_t)-1;
}
and the COBOL call looked like this:
json generate result from grp
call "CNV" using by reference result,
by value length of result,
by reference convertedres,
by value length of convertedres,
returning cres
and cres is a PIC S9(9) COMP data item which will have a non-zero value of the conversion succeeded.
Again, I apologize for not first asking if C++ is a possibility. (Or even C. The code could be easily done in C.) Also, the result is not quite perfect owing to the JSON GENERATE result being zero filled.
I have written a small code to fetch info of 'n' number of students.
But after running the program, I'm getting a segfault. Please find the code below.
struct students
{
char name[20];
int age;
int id;
}student[100];
int main()
{
int count;
int no_students;
printf("Enter no of students");
scanf("%d",&no_students);
for (count = 1 ; count <= no_students ; count++)
{
printf("Enter the details for student%d\n",count);
printf("Name:");
scanf("%s",student[count].name);
printf("Age:");
scanf("%d",student[count].age);
printf("ID:");
scanf("%d",student[count].id);
}
return 0;
}
root#debian:/home/renga/C_code# ./nike
Enter no of students3
Enter the details for student1
Name:renga
Age:12
Segmentation fault
Got the problem, missed reference operator in scanf.
printf("Age:");
scanf("%d",&(student[count].age));
printf("ID:");
scanf("%d",&(student[count].id));
This may not be the right group to ask C language help/syntax question. I would recommend reading a good C programming language book ex: http://www.cprogramming.com/tutorial/c-tutorial.html
Having said that, the problem is, you are trying to read your int data type (age & id) values incorrectly. You need to read the values into the address of the variables as shown below...
scanf("%d",&student[count].age);
I have a question regarding the converting string to intvalue. My question and issue is in case if I have string called "001223" I am getting 1223 as intvalue. But I want to get the 001223 as final int value. Please let me know if my question is not clear. Thanks for your time
There is no difference in value between the numbers 001223 , 1223, 2446/2 or 1223.000. They all refer to the same number.
If you want to keep leading zeroes, then you need to either keep it as a string or maintain another piece of information so it can be rebuilt later, basically the number of zeroes at the front, such as:
struct sNumWithLeadingZeros {
size_t zeroCount;
unsigned int actualValue;
};
I'd probably suggest the former (keeping it as a string) since that's likely to be less effort.
"Leading zeros" are to do with the textual representation of an integer, when stored as integer values in a computer the leading zeros do not exist.
However, if what you want to do is display the number with the same number of digits it had before being converted from text then: if the string contains only the digits of the number, e.g. you have #"001223" then you can take the length of this string to determine the number of digits. Later when converting the number back to string format you can use a formatted conversion, e.g. stringWithFormat:, and a format specifier which specifies the required number of digits. You'll need to read up on formats in the documentation, but here is an example:
NSString *input = #"001223";
int x = [input intValue];
int digits = (int)input.length;
NSString *output = [NSString stringWithFormat:#"%0*d", digits, x];
The value of output will be the same as input. The format broken down is: 0 - leading zeros; * use a dynamic field with, will use the value of digits; d - int.
HTH
One cannot prefix leading 0s in int data type. But if you see 0 prefix then the number is octal not decimal. Octal value can be created by changing base. For this you can use wrapper class like Integer.
But if one wants leading 0s for displaying data then he/she can use following code
public class Sample
{
public static void main(final String[] argv)
{
System.out.printf("%06d", 1223);
System.out.println();
}
}
I am trying to make a small application that prints the content of a number of consecutive memory locations. As an indication of where the program is in the memory, I am printing the memory location of the main function and of a dummy variable.
In a first column, I want to print the address. In a second column, I want the contents of this address and the content of the 9 addresses behind it. In a third column, I want to print the byte value as a char, if it is printable. If it is not printable, I want to print a dot. In the rows under the first, I do exactly the same.
At startup, a number of values to print can be entered. If a positive value is entered, the addresses will increment, if a negative value is entered, the addresses will decrement.
To quickly see where I want to get to, you could run the code and enter for example 20 bytes to dump and use the address of the dummy as a starting address.
So far, my code only works for positive values. When I enter a negative number, I get a segmentation fault, but I can't figure out why. I tried to find the error in Valgrind, without success.
Some help would be greatly appreciated!
Here's my code:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define NUMBER_OF_BYTES 10
void showAddresses(void);
void printMemory(void);
void getDumpSize(void);
char* dummy; // dummy is een pointer naar een int
signed int dumpSize; // how many bytes have to be dumped?
signed int upDown; // do I need to go up or down?
int main(void)
{
dummy = (char *) malloc (sizeof(unsigned char));// memory allocation
showAddresses(); // prints the main function address and a variable address
getDumpSize(); //
printMemory(); //
free(dummy); // free memory
return 0; // end the main function
}
void showAddresses(void)
{
printf("Main function address is %p \n", main);
printf("Dummy variable address is %p \n",(void*)dummy);
}
void getDumpSize(void)
{
printf("Enter number of bytes to dump <negative or positive>:");
scanf("%d",&dumpSize);
if(dumpSize<0)
{
upDown = -1; // count down
printf("upDown was set to -1\n");
}
else
{
upDown = 1; // count up
printf("upDown was set to +1\n");
}
}
void printMemory(void)
{
int input;
printf("Enter the address:");
scanf("%x", &input); // enter the input
printf("Address \tBytes \t\t\t\tChars \n"); // print the table header
printf("--------- \t----------------------------- \t---------- ");
int i;
unsigned char* address; //
for(i=0;i<abs(dumpSize);i++)
{
address = (unsigned char*) (input+(i*upDown)); // make the address to print
if( (i%NUMBER_OF_BYTES) == 0) // show the address every 'i*NUMBER_OF_BYTES' times
{
printf("\n%p \t", (void*) address);
}
printf("%02x ", *address); // print as a 2 number hex and use zero padding if needed
if( (i%NUMBER_OF_BYTES) == (NUMBER_OF_BYTES-1) )// print the char list for every value (if printable)
{
printf("\t");
int j;
for(j=(NUMBER_OF_BYTES-1);j>=0;j--)
{
address = (unsigned char*) (input+(i*upDown)-j);
if(isprint(*address)==0)// print a dot if the byte value is not printable
{
printf(".");
}
else
{
printf("%c",*address); // print the byte value as a char, if printable
}
}
}
}
}
Your segmentation fault is likely coming from attempting to access memory outside of your program's scope. The address in "dummy" is the first malloc from your program, so it represents (possibly) the first "area" of memory available to your program. Going up from there might be keeping you in program space (hence the lack of seg fault), but going backward might be driving you into a restricted memory area. Out of curiosity: what is the memory address returned for your "dummy" malloc? Is the number even large enough to go backward without going negative? (I'm wondering if your program sees the true system memory map or a paged map that is already sand-boxed for programs.)
I'm writing a program using linked list (such a nightmare).
Anyway, the purpose of the program is to enter 8 characters and have the program print the characters back out to you and also print the characters back out in reverse order, using linked lists of course.
I got this so far. There's a lot wrong with it (i think).
Problems are
When asking for characters from the user it should read in the amount of characters automatically without having to ask for how many characters
Also, when it it compiles it prints gibberish to the screen, for example I just ran it and it printed
¿r
(àõ($ê¿¿
a¿r
(àõ($ê¿¿
¿r
(àõ($ê¿¿
b¿r
(àõ($ê¿¿
Lots of help needed here. It would be so much appreciated!
Code of course
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define strsize 30
typedef struct member
{
int number;
char fname[strsize];
struct member *next;
}
RECORD;
RECORD* insert (RECORD *it);
RECORD* print(RECORD *it, int j);
int main (void)
{
int i, result;
RECORD *head, *p;
head=NULL;
printf("Enter the number of characters: ");
scanf("%d", &result);
for (i=1; i<=result; i++)
head=insert (head);
print (head, result);
return 0;
}
RECORD* insert (RECORD *it)
{
RECORD *cur, *q;
int num;
char junk;
char first[strsize];
printf("Enter a character:");
scanf("%c", &first);
cur=(RECORD *) malloc(sizeof(RECORD));
strcpy(cur->fname, first);
cur->next=NULL;
if (it==NULL)
it=cur;
else
{
q=it;
while (q->next!=NULL)
q=q->next;
q->next=cur;
}
return (it);
}
RECORD* print(RECORD *it, int j)
{
RECORD *cur;
cur=it;
int i;
for(i=1;i<=j;i++)
{
printf("%s \n", cur->fname);
cur=cur->next;
}
return;
}
You have:
in insert:
char first[strsize];
scanf("%c", &first); /* note the %c */
strcpy(cur->fname, first);
in print
printf("%s \n", cur->fname);
You should have %s instead of %c and therefore change &format to format in the argument list, as format itself represents the address of the location the string is to be stored.
So the scanf call should be like below
scanf("%s", first);
Another thing. If you have specified a return type in the print function then you should return something, or make it return nothing (declare return type as void). This will not pose any problem in this case although.
Read the warning messages which the compiler throws to you and you would see the compiler actually had answered your questions.
You need to do some redesigns i think. For example to traverse the linked list you do not need to counter 'j'. you can detect the list termination by inspecting if the next link is NULL or not.
Your question was to print the characters or strings in reverse, so you need to write some other print function than what you have wrote.