What happens if you free a pointer twice?
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Write a Program to print this triangle: * ** * **** * ****** * ******** * ********** use two nested loops.
12 Answers MIT, TCS,
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If I want to initialize the array like. int a[5] = {0}; then it gives me all element 0. but if i give int a[5] = {5}; then 5 0 0 0 0 is ans. what will I do for all element 5 5 5 5 5 in a single statement???
void main() { int s[4][2]={ {1234,56},{1212,33},{1434,80},{1312,78} }; int (*p)[2]; int i,j,*pint; for(i=0;i<=3;i++) { p=&s[i]; pint=p; printf("\n"); for(j=0;j<=1;j++) printf("%d",*(pint+j)); } } while running this program it shows a warning-suspicious pointer conversion ie pint=p; my que is why should we assign the value of p to pint again.why cant we use it directly as *(p+j)..but if i use like tat the o/p is garbage value..
1,1,5,17,61,217,?,?.
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