1. Assuming the following structure begins at address 6000, show the output. Note that doubles take 8 bytes struct double dGPA; char szName [20]; char szMajor[4]; student -(3.5, "Bill Board", "CS"I; int i 2; char *pCh; Output: printf("A. Addr of dGPA-%pln" , &student.dGPA); printf( B. Addr of szName pn" , student.szName) printf("C. Addr of szMajor-pln" , student.szMajor); p Ch-&student.szN ame[i]; printf("D. Addr of szName[i]-pn" pCh) PCh++ printf("E. Addr of pCh-%p ", pCh); printf("F. pCh); String at pCh-%s