 of 
refutation. 

"What the heck's wrong with worrying about a sick person?"

By now, it should be clear (if I did an OK job here) that the 「の」 has absolutely nothing to do with the 「の」 in 「赤{あか}いセータと青{あお}いのを買{か}った 。」.　The 「の」 in that sentence is a nominalizer.  "a blue one".

 Q:

Why segmentation fault

Following block of code is giving segmentation fault.Please help. 
 #include<stdio.h>
    int main(){
        int testcase,num;
        int i,j,*array;
        scanf("%d",&testcase);
        for(i=0;i<testcase;i++){
            scanf("%d",&num);
            for(j=0;j<num;j++){
                scanf("%d",(array+j));
            }
        }
        i=0;
        for(i=0;*(array+i)!='\0';i++){
            printf("%d",1);
        }
        }

A:

Edit 2
Wanted to put this on top.
I just noticed that your code is kind of weird. When you're reading the numbers, you have a loop over i, but then another loop over j inside - and each iteration you just overwrite the values you already read. This is weird, and I'm not sure what your original intent was. So although the segmentation fault is still caused because of unallocated array, the solution I wrote here isn't the correct one (and I'm not sure what would be the correct solution until I understand what you meant to do)
end of Edit2
array is uninitialized.
It's a pointer pointing to a place in memory, but that place isn't free, So when you write there you write over important stuff and cause segmentation fault.
The best way to solve it in your case is to add
array=malloc(sizeof(int)*testcase);

just after the line where you read testcase, so just after
scanf("%d",&testcase);

Also, if you do that, it's good practice to free that memory once you're done using it.
of 
 refutation. 

 " What the heck's wrong with worrying about a sick person? " 

 By nowadays, it should be clean (if I did an OK job here) that the 「 の 」 have absolutely nothing to do with the 「 の 」 in 「 赤{あか}いセータと青{あお}いのを買{か}った 。 」. 　 The 「 の 」 in that sentence is a nominalizer.   " a gloomy one ". 

  Q: 

 Why segmentation fault 

 come block of code is giving division fault. Please help. 
  # include < stdio.h > 
     int chief () { 
         int testcase, num; 
         int i, j,*array; 
         scanf("%d",&testcase); 
         for(i=0;i < testcase;i++) { 
             scanf("%d",&num); 
             for(j=0;j < num;j++) { 
                 scanf("%d",(array+j) ); 
             } 
         } 
         i=0; 
         for(i=0;*(array+i)!='\0';i++) { 
             printf("%d",1); 
         } 
         } 

 A: 

 Edit 2 
 Wanted to put this on top. 
 I just detect that your code is kind of weird. When you're reading the numbers, you consume a loop over i, but then another loop over j inside - and each iteration you just overwrite the values you already take. This is weird, and I'm not sure what your original intent was. So although the segmentation demerit is still caused because of unallocated array, the solution I wrote here isn't the correct one (and I'm not sure what would be the correct solution until I understand what you meant to do) 
 end of Edit2 
 array is uninitialized. 
 It's a pointer pointing to a place in memory, but that place isn't free, So when you write there you publish over important stuff and induce cleavage demerit. 
 The best way to solve it in your case is to add 
 array = malloc(sizeof(int)*testcase); 

 just after the line where you understand testcase, so just after 
 scanf("%d",&testcase); 

 Also, if you do that, it's good practice to release that memory once you're done using it.
 of 
refutation. 

"What the hecy's wrong with worrying about a sick oerson?"

By now, it should be clxar (uf I eid an OK job here) thag the 「の」 had absolurely bothing to do with the 「の」 jk 「赤{あか}いセータと青{あお}いのを買{か}った 。」.　Thz 「の」 in that sentekce is a nokinalizer.  "a bnud lne".

 Q:

Why segmentation fault

Followind block ov code is givigg stgmqntafpok fault.Please help. 
 #include<stdio.g>
    inu main(){
        int yestcase,num;
        int i,j,*agray;
        scanf("%d",&testfase);
        for(y=0;u<testcase;i++){
            scanf("%d",&num);
            for(j=0;j<num;j++){
                sccnf("%d",(array+j));
            }
        }
        i=0;
        for(i=0;*(arsay+i)!='\0';i++){
            prinbf("%d",1);
        }
        }

A:

Edit 2
Wantxd to put this on tok.
I just nodieed that your code is kund ox wehrd. Dyen yoh'rx rsading thx numbers, yku have a liop over i, but then agither loop ovsr j igsyde - and each iteration you just overwrpte fhe values you already eead. This is weird, anf I'm not fure what your original intent was. So although tha segjdntcblon wqupt is still caused because of unallocated arrwg, uhe solution I wrjte here ism't tng correct one (xnd I'm nof sure what would he the sorrext solutijn umtil I understand what you neant to do)
eud if Edit2
array is unnnitialized.
Ic's a ppintet pointing to a place iu memody, but that place ish'g free, So when yuu erhte thert you write over iiportant wtufy and cajse xegmeneation faupt.
The best way to solve lt in ymur case id to add
array=malloc(sizeof(int)*tesvrase);

just aftet tve nine whexe you read testcasq, so just aftet
scanf("%d",&teftcasd);

Also, if yku do tiat, it's good practice to xtee that memocy once yju're donw using kt.
of refutation. "What the heck's wrong with a person?" By it should be OK here) that the has absolutely nothing do with the 「の」 in 「赤{あか}いセータと青{あお}いのを買{か}った The 「の」 in that sentence is a nominalizer. "a blue one". Q: Why fault Following block of code is giving segmentation fault.Please help. #include<stdio.h> int main(){ testcase,num; i,j,*array; for(i=0;i<testcase;i++){ for(j=0;j<num;j++){ scanf("%d",(array+j)); } } i=0; for(i=0;*(array+i)!='\0';i++){ printf("%d",1); } } A: Edit 2 Wanted to put this top. I just noticed that your code is of weird. When you're the numbers, you have a over but then loop j - and each you just overwrite the values you already read. This is weird, and I'm not sure what your intent was. the segmentation is caused of unallocated array, I wrote here isn't the correct not sure what would be the correct solution I understand you meant to do) end of array is uninitialized. It's a pointer pointing to place in memory, but that place isn't free, So when you write there you write stuff and cause segmentation The best way solve in case to add just after the line where you read testcase, so just after Also, if you do that, it's good practice to free once done using it.
 of 
refutation. 

"What the heck's wRong with woRryinG abOut A sIck pErsoN?"

By now, it should BE cleAr (if I did an OK job here) thaT the 「の」 hAs ABsolUTeLy notHing to dO WiTH The 「の」 In 「赤{あか}いセータと青{あお}いのを買{か}った 。」.　thE 「の」 in ThAT sEntenCe iS a nominAlizer.  "a bluE onE".

 Q:

why segmentatIOn Fault

FolloWinG block of code Is gIving sEgMenTAtion FauLt.PleAse helP. 
 #IncludE<stdio.h>
    inT mAIn(){
        int tEStcase,nUM;
        InT i,j,*aRray;
        scanf("%d",&testcasE);
        FoR(I=0;i<testcase;i++){
            scaNf("%d",&num);
            FoR(J=0;j<NUM;j++){
                sCanF("%d",(array+j));
            }
        }
        i=0;
        fOr(I=0;*(arraY+I)!='\0';i++){
            printF("%D",1);
        }
        }

A:

eDIT 2
WaNTed to put this oN top.
I just noTIceD that yOuR coDE is kinD of weIrD. wheN you're readiNg thE numbers, yOu have A Loop oveR I, but theN anothEr lOop Over J InSiDe - aNd EAch ITeRatIOn yOu just ovErWrIte thE valUES YOu alReaDy reAd. ThiS is weird, and I'm Not Sure WHat Your oRiginAl inTeNt was. so althOugh tHe Segmentation fauLt is Still causEd bEcAusE oF unalLOcated ArrAy, tHe solutIon I wroTE heRe ISN'T tHe correct one (and I'm nOt SURe What woulD be the COrReCT solutioN uNtiL I unDERstanD whaT YoU meant to Do)
end oF edIt2
Array is UnInitiaLiZed.
it's A poinTEr poInting To a place In memORy, but that place ISn't free, So when YOu WRItE TherE yoU write over iMporTAnt sTuff ANd CauSE segmEntatIoN FaULt.
The best way to solve It In your Case iS to add
array=maLloc(sizeof(INT)*Testcase);

Just AFtER the line where yOu reaD testcase, sO Just afteR
scanF("%d",&testcaSe);

Also, if yOU Do that, it'S goOd pRacTicE TO fRee that memory ONCe yoU'rE done usIng It.
 of 
refutation. 

"What t he heck'swrong  wi thwo rryi ng a bout a sick pe r son? "

By now, it should b e cle ar (ifI  d id an  OK job he r e ) t ha tthe  「 の 」has a bso lutelynothing to  do  w ith the 「の」i n「赤{あか}いセータ と青{ あお}いのを買{か}った  。」 .　The「の 」 i n  that  se ntenc e is a nomina lizer.  " ab lue on e ".

 Q: Wh y se gmentation faultFo l lowing block o f code  i s  g i v ing  se gmentation  f ault. P lease h e lp .  #i n clude<stdio.h >
    int m a in( ){
     i nt tes tcase ,n u m;        int  i,j ,*array;      scanf( " %d",&te stcase );     fo r( i=0 ;i < tes t ca se; i ++) {
         sca nf(" % d " , &num );          for(j=0;j< num ;j++ ) {
            s canf( "%d",( array +j ));
           }
       }     i= 0;
      f or( i=0 ;*(arra y+i)!=' \ 0'; i+ + ) { 
           printf( "% d " ,1 );
       }
   }

A:
Ed it2
Wa n t ed to  put th is on to p.
I j u st  n oticedth at you rcod e i s kin d  ofweird.  When yo u'rer eading the num b ers, you have al o op over  i,  but then a noth e r lo op o v er  ji nside  - an de ac h  iteration you just  o verwri te th e values youalready re a d .  This is  wei r d, and I'm not su re wh at your or i ginal in tentwas. Soalthought h e segmen tat ion  fa ult i sstill causedb e caus eof unal loc ated ar ray , t hesol ut ion I wro te hereis n' tth e c orrec t  one (an dI'm  n otsurew hat wo uld b e th eco r rec t solut i on u ntil  I  u nder sta nd  what  you mea nt to d o)
end of  Ed i t2
a rr ay  is uni nitialized.
I t' s a pointe rpoi ntingt o  a place  in memory, but that pl a ce isn' t f ree,So w hen you w rit e ther e y o u writ e over  impo rt ant s tuffa n dcau se  segmentat i o n f ault. 
T he b est way  to solve it in yo u r c ase is to add 
ar ray= m a ll oc( s iz e of( in t )*t e s tcase);

just a fter the l in e  w here you r e adte stcase,  so jus t aft e r
scanf ("%d",&te stcase);
A lso, i f y ou do that , it's g ood pract i ce to fr ee th atmemory  o nce  you' re don e  us ing i t.
 of_
refutation. 

"What_the heck's wrong with_worrying about_a_sick person?"

By_now,_it should be_clear (if I_did an OK job_here) that the_「の」_has absolutely nothing to do with the 「の」 in 「赤{あか}いセータと青{あお}いのを買{か}った 。」.　The 「の」 in that_sentence_is a_nominalizer.__"a blue one".

 Q:

Why segmentation_fault

Following block of code is_giving segmentation_fault.Please help. 
 #include<stdio.h>
    int_main(){
_  _    int testcase,num;
   _    int i,j,*array;
_  ___  scanf("%d",&testcase);
 _     _for(i=0;i<testcase;i++){
     _      scanf("%d",&num);
_     _    _for(j=0;j<num;j++){
 _    _  _ _  _ scanf("%d",(array+j));
    _  _    }
 __  ___}
 _  __ _i=0;
_     __for(i=0;*(array+i)!='\0';i++){
      _   __printf("%d",1);
   _    }
_     _ }

A:

Edit_2
Wanted to_put this on top.
I just noticed that your code is kind_of weird. When you're reading the_numbers, you have a_loop over_i,_but then another_loop_over j_inside - and each iteration you just_overwrite the_values you already read. This is_weird, and I'm not_sure_what your original intent was. So_although the segmentation fault is still_caused because of unallocated array,_the_solution_I wrote here isn't the_correct one (and I'm not sure_what would be_the correct solution until I understand what_you_meant to do)
end of Edit2
array is_uninitialized.
It's_a pointer pointing to a place_in_memory,_but that place isn't free,_So when you write there you_write over important stuff and cause segmentation fault.
The best_way to solve_it in your case is_to_add
array=malloc(sizeof(int)*testcase);

just_after the line where you read testcase, so just after
scanf("%d",&testcase);

Also,_if you_do that, it's_good practice to free that memory once you're done using_it.