Question

Q3, What is the output of the program shown below and explain why. #include #include <stdio.h> <string.h> int main(void) ( char str[ 39764 int N strlen(str) int nunj for (int i-0; i t Nǐ.de+ printf(") num (str[] 'e); I/ digits range in the ASCII code is 48-57 if (nue > e) f 9 printf( Xdin, num)): else ( for (int j e; 1<N 1; j printf("- ) printf(In")

can u please solve it in c++ format,,,, general format which is #include<stdio.h> .......printf and scanf format

dont worry about the answers i have down

Answer 1

Hi, it is not clear what are you trying to ask. Your code is given in C language and not in C++. Also you have asked to solve it in C++ but at the same time you are specifying that general format for that is printf, scanf which is not true, it is for C language and not for C++. Do you want the above code to be converted to C++ from C or do you want its output after compiling along with explanation.

Anyways, am doing all of these, but sorry if you meant something else. Pls don't rate bad for this.

Code and its output in C -

#include<stdio.h>
#include<string.h>

int main(void)
{
char str[]= "39704";
int N= strlen(str); /* here N = 5, since length of str[] is 5 */
int num;

/* trasversing along full string str[] */

for(int i=0; i<N; i++)
{
for(int j=0; j<i; j++) /* this for loop creates 0 space in first row, 1 space in next row, and so on */
printf(" ");  
num= (str[i]- '0'); /* num gets integer value of i th str[] element (ASCII value concept is prerequisite for this), for eg, '3' - '0' gives 3, bcos, if we directly see str[0], we will get str[0] as 51(which is ASCII code of 3), and '0' is 48, so we do '3' - '0' and we get 3 (actually 51-48 =3 ) similarly, '9' - '0' gives 9 (actually 57-48 = 9 )*/
if(num > 0)
{
for(int j=0; j< N-1-i; j++) /* this loop multiplies 10 to num till N-1 times */
num *=10;
printf("%d ", num); /* prints value of num till now */
}
else /* enters here if num is less than OR equals to zero*/
{
for(int j=0; j< N-i; j++)
printf("-");
printf(" ");
}
}
}

Code output-

I am also writing code in C++ -

#include <iostream>
#include<cstring>
using namespace std;

int main(void)
{
char str[]= "39704";
int N = strlen(str);   /* here N = 5, since length of str[] is 5 */
int num;

  /* trasversing along full string str[] */

for(int i=0; i<N; i++)
{
for(int j=0; j<i; j++) /* this for loop creates 0 space in first row, 1 space in next row, and so on */
cout<<" ";
num= (str[i]- '0'); /* num gets integer value of i th str[] element (ASCII value concept is prerequisite for this), for eg, '3' - '0' gives 3, similarly, '9' - '0' gives 9 */
if(num > 0)
{
for(int j=0; j< N-1-i; j++) /* this loop multiplies 10 to num till N-1 times */
num *=10;
cout<<num<<endl; /* prints value of num till now */
}
else    /* enters here if num is less than OR equals to zero*/
{
for(int j=0; j< N-i; j++)
cout<<"-";
cout<<endl;
}
}
}