Unable to obtain Matrix in C [closed] - c

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Following is the code to obtain matrix like 11 12 13 etc... I am not getting desired results since long. Help me with the solution.
#include<stdio.h>
int main() {
int i,j;
for(i=1;i<5;i++){
for(j=1;j<5;j++){
printf("%d %d",i,j);
}
printf("\n");
}
fgetc(stdin);
}

Probably because you have a space between the digits:
printf("%d %d",i,j);
Shouldn't it be:
printf("%d%d ",i,j);
for your desired output?

Shouldn't printf("%d %d",i,j); actually be printf("%d%d ",i,j);?

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I am a beginer at programming,I wrote a code which gives perfect output but online judge didn't accepted.Why it's happen?(Codeforces problem 69A. ) [closed]

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#include<stdio.h>
int main(){
int a,b[101],i,j,sum=0;
scanf("%d",&a);
for(i=0;i<a;i++){
for(j=0;j<3;j++){
scanf("%d",&b[j]);
sum=sum+b[j];
}
}
if(sum==0) printf("YES\n");
else printf("NO\n");
return 0;
}
outputs are showed perfectly but online judge didn't accept.
online judge message"wrong answer on test 81".
link of problem"http://codeforces.com/problemset/problem/69/A"
The problem is asking to check if the sum of all vectors is zero.
Sum of vectors is element-wise addition of vectors, not simple summation of all elements.
Your program will fail in, for example, this case:
1
1 0 -1

Incrementing constant variable [closed]

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#include <stdio.h>
#define number 0
main()
{
int i;
for (i = 0; i < 5; i++)
{
number++;
}
printf("Number is: %d", number);
}
No, 0 is substitued for number by the preprocessor before the compiler gets to work.
The compiler issues a diagnostic when it sees 0++;

How to calculate Determinant Matrix 2x2 -C [closed]

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So I need to calculate the determinant matrix on 2x2 matrix, simple one..
If you can, please write the code, because i don't understand like in the other thread, it's like i'm reading an English Literature + Code..
#include<stdio.h>
int main(){
int x[2][2]={{1,2},{3,4}};
int i,j,d;
for(i=0;i<=1;i++){
for(j=0;j<=1;j++){
printf(" %d ", x[i][j]);
}
printf(" \n");
}
return 0;
}
Thank You!
You don't need a loop to calculate the determinant of a 2x2 matrix, simply use:
int result = (x[0][0] * x[1][1]) - (x[0][1] * x[1][0]);

Why does this print 1? [closed]

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#include <stdio.h>
int main()
{
int var=0;
for(; var++; printf("%d",var));
printf("%d", var);
}
Please explain to me this C code. How is the output 1?
You might be confused because of the wrong code indentation. Your code is:
for(; var++; printf("%d",var))
;
printf("%d", var);
So you always get the output of the second printf. As var is initialized to 0 and var++ (the for-condition) is always executed, you end up with var==1.

Thread stopped...access violation at 0x6: read of address 0x6 [closed]

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I am getting an error saying: Thread stopped.....access violation at 0x6:read of address 0x6.
However my code doesn't have such address.
#include<stdio.h>
#include<conio.h>
void main ()
{
int i,j,A[4][5],c=0;
for(j=0;j<5;j++)
for(i=0;j<4;i++)
{
A[i][j]=c;
c++;
}
for(i=0;i<4;i++)
{
for(j=0;j<5;j++)
printf("%d\t",A[i][j]);
printf("\n");
}
getch();
}
I was asked to enter a double array column by column instead of row by row, and I want to verify my method so I tried to print it out but had no luck doing so.
for(i=0;j<4;i++)
^^
Shouldn't this be:
for(i=0;i<4;i++)
j<4 is always true once it enters inner the loop for the first time and is the reason for memory access violation.

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