This us what turned up to be the solution to my problem; http://www.cplusplus.com/forum/beginner/144062/

https://paste.debian.net/1178652/ I've written this function to turn a char* Buffer into a char** array of lines, but it doesn't work. Can you spot the error?

your last line might not contain a newline in which strchr will return NULL

Hey. I am beginner and do not even know the alphabets of programming. Can somebody help md learn the basics.

Hey. I am beginner and do not even know the alphabets of programming. Can somebody help md learn the basics.

Program calculates all prime numbers less than x

Help

Program calculates all prime numbers less than x

hi guys, anyone around? i've got a doubt

I cant grasp the idea of why arrays are non-modifiable lvalues... I've read some documentation online and I know it has something to do with C predecessor B. I didnt get if the name of the array is just a constant number or what. I get what lvalues/rvalues are, I get how pointers work but I dont understand why in the name of god I cant do smth like this: int x[10]; int y[10]; x = y;

x is int*

I dont, Im just trying to understand why arrays are non-modifiable lvalues, that just the example I came up with

yeah but basically the name of the array is the address of the 1st element

I mean I get it but wanna know the WHY

So they more of int * const

Different countries use different coin denominations. For example, Bangladesh uses 1, 2, and 5. A desirable property of coin denominations is to have each coin at least twice the amount of its previous coin in sorted order. For example, the Bangladesh coin denominations have this property, but the coin denominations {1, 5, 6} do not (6 is not at least twice 5). The Problem: Given the coin denominations, you are to determine if the set has the above property. Input Format The first input line contains a positive integer, n, indicating the number of denomination sets to check. The sets are on the following n input lines, one set per line. Each set starts with an integer d , which is a count of various coin amounts in the set; this is followed by d distinct positive integers (each less than 1,000) giving each coin amount (assume the coin amounts are given in increasing order). Constraints (1 ≤ d ≤ 10) Output Format At the beginning of each test case, output “Denominations: v” where v is the input values. Then, on the next output line, print a message indicating whether or not the set has the above property. Leave a blank line after the output for each test case. Follow the format illustrated in Sample Output. Sample Input 0 2 4 1 5 10 25 3 1 5 6 Sample Output 0 Denominations: 1 5 10 25 Good coin denominations! Denominations: 1 5 6 Bad coin denominations! (can anyone plz solve this c program)

Hello, i'm beginner I want to ask, an example this array is: char vowel[5]="aiueo" char konsonan[21]="bcdfghjklmnpqrstvwxyz" these array can be collected into one array??

char a[N]; char b[M]; // create an array that could fit contains of both char newarr[(sizeof(a) + sizeof(b) + 1)]; // +1 for '\0' // copy contents of these arrays into the new one memcpy(newarr, a, sizeof(a)); memcpy(newarr + sizeof(a), b, sizeof(b)); // add '\0' so printing it won't make your pc on fire newarr[sizeof(newarr) - 1] = '\0';

Example! char A[] ="string"; char B[11] ="string"; char C[] ={'s', 't', 'r', 'i', 'n', 'g', '\0'}; char D[11] ={'s', 't', 'r', 'i', 'n', 'g', '\0'}; For A & B '\0' NULL is added automatically, but in other case it needs to be added explicitly. —-now your case!! You are using +1 or should it be +2?!!😄

Example! char A[] ="string"; char B[11] ="string"; char C[] ={'s', 't', 'r', 'i', 'n', 'g', '\0'}; char D[11] ={'s', 't', 'r', 'i', 'n', 'g', '\0'}; For A & B '\0' NULL is added automatically, but in other case it needs to be added explicitly. —-now your case!! You are using +1 or should it be +2?!!😄

Also this string doesn't have a terminator, because he explicitly wrote a size without it char vowel[5]="aiueo"

Try running the above code with different array size & let know what output you get & and why?? —-[1] modify & run!! char A[6] = "hello\n"; char B[7] = "hello\0"; char C[7] = "hello"; —-[2] modify & run!! char A[7] = "hello\n"; char B[5] = "hello\0"; char C[5] = "hello"; —-[3] modify & run!! char A[7] = "hello\n"; char B[6] = "hello\0"; char C[5] = "hello"; 😄

So to resume: char C[5] = "hello"; does not have a null terminator

If you say there is not NULL termination! And printf() is misbehaving try cout!! — include <iostream> #include <string.h> using namespace std; int main(void) { char A[7] = "hello\n"; char B[7] = "hello\0"; char C[7] = "hello"; char D[(sizeof(B) + sizeof(C) + 1)]; cout << "\n---A"; cout << "Array[size]:" <<sizeof(A) << " Str:"<<A <<"|len:" <<strlen(A); cout << "\n---B"; cout << "Array[size]:" <<sizeof(B) << " Str:"<<B <<"|len:" <<strlen(B); cout << "\n---C"; cout << "Array[size]:" <<sizeof(C) << " Str:"<<C <<"|len:" <<strlen(C); return 0; } —-[1] modify & compile char A[7] = "hello\n"; char B[6] = "hello\0"; char C[7] = "hello"; —-[2 modify & compile char A[7] = "hello\n"; char B[7] = "hello\0"; char C[5] = "hello"; 😄

Who would have thought

Try this!! & find how many NULL do you see in Bx & Cx.. include <iostream> #include <string.h> using namespace std; int main(void) { char A[7] = "hello\n"; char B[7] = "hello\0"; char C[7] = "hello"; cout << "\n---A"; cout << "Array[size]:" <<sizeof(A) << " Str:"<<A <<"|len:" <<strlen(A); cout << "\n---B"; cout << "Array[size]:" <<sizeof(B) << " Str:"<<B <<"|len:" <<strlen(B); cout << "\n---C"; cout << "Array[size]:" <<sizeof(C) << " Str:"<<C <<"|len:" <<strlen(C); cout <<"\n---BX" <<endl; for(int i=0; i<(sizeof(B)); i++) { cout <<" B-Str:" <<B[i] <<endl; if(B[i] == '\0') cout <<"shout NULL" <<endl; } cout <<"\n---CX" <<endl; for(int i=0; i<(sizeof(C)); i++) { cout <<" C-Str:" <<C[i] <<endl; if(C[i] == '\0') cout <<"shout NULL" <<endl; } return 0; } 😄 try changing size too..

Try this!! & find how many NULL do you see in Bx & Cx.. include <iostream> #include <string.h> using namespace std; int main(void) { char A[7] = "hello\n"; char B[7] = "hello\0"; char C[7] = "hello"; cout << "\n---A"; cout << "Array[size]:" <<sizeof(A) << " Str:"<<A <<"|len:" <<strlen(A); cout << "\n---B"; cout << "Array[size]:" <<sizeof(B) << " Str:"<<B <<"|len:" <<strlen(B); cout << "\n---C"; cout << "Array[size]:" <<sizeof(C) << " Str:"<<C <<"|len:" <<strlen(C); cout <<"\n---BX" <<endl; for(int i=0; i<(sizeof(B)); i++) { cout <<" B-Str:" <<B[i] <<endl; if(B[i] == '\0') cout <<"shout NULL" <<endl; } cout <<"\n---CX" <<endl; for(int i=0; i<(sizeof(C)); i++) { cout <<" C-Str:" <<C[i] <<endl; if(C[i] == '\0') cout <<"shout NULL" <<endl; } return 0; } 😄 try changing size too..

Are past terminator values guaranteed to be zero initialized in the standard?

Try this!! & find how many NULL do you see in Bx & Cx.. include <iostream> #include <string.h> using namespace std; int main(void) { char A[7] = "hello\n"; char B[7] = "hello\0"; char C[7] = "hello"; cout << "\n---A"; cout << "Array[size]:" <<sizeof(A) << " Str:"<<A <<"|len:" <<strlen(A); cout << "\n---B"; cout << "Array[size]:" <<sizeof(B) << " Str:"<<B <<"|len:" <<strlen(B); cout << "\n---C"; cout << "Array[size]:" <<sizeof(C) << " Str:"<<C <<"|len:" <<strlen(C); cout <<"\n---BX" <<endl; for(int i=0; i<(sizeof(B)); i++) { cout <<" B-Str:" <<B[i] <<endl; if(B[i] == '\0') cout <<"shout NULL" <<endl; } cout <<"\n---CX" <<endl; for(int i=0; i<(sizeof(C)); i++) { cout <<" C-Str:" <<C[i] <<endl; if(C[i] == '\0') cout <<"shout NULL" <<endl; } return 0; } 😄 try changing size too..