出二维数组周边元素之和作为函数值返回
第一个for循环用于计算矩阵最上一行和最下一行的总和，第二个for循环则计算除两头元素以外的最左一列和最右一列的元素的和。最后，将这两个求和结果相加得到周边元素的和。
〔源程序〕
#include
#include
#define M 4
#define N 5
int fun(int a[M][N])
{int sum=0,i;
for(i=0;i sum+=a[0][i]+a[M-1][i]; for(i=1;i sum+=a[i][0]+a[i][N-1]; return sum ; } main() { int aa[M][N]={{1,3,5,7,9}, {2,9,9,9,4}, {6,9,9,9,8}, {1,3,5,7,0}}; int i,j,y; clrscr(); printf(＂The original data is :\n＂); for(i=0;i {for(j=0;j printf(＂\n＂); } y=fun(aa); printf(＂\nThe sum: %d\n＂,y); printf(＂\n＂); } private function Anyl(A()() AS INTEGER,byval m as integer,byval n as integer) as long tol=sum_side(A,m,n) print tol if m=n then sd=Diag(A,m) sid=InDiag(A,m) print sd print sid else print ＂m≠n＂ end if end function private function Sum_side(A()() AS INTEGER,byval m_side as integer,byval n_side as integer) as long dim sum as long sum=0 n_side=n_side-1 for i=0 to n step 1 sum=sum+A(0)(i) next i m=m-2 for i=1 to m step 1 sum=sum+A(i)(0) sum=sum+A(i)(n) next i m=m+1 for i=0 to n step 1 sum=sum+A(m)(i) next i sum_side=sum end function private function Diag(A()() AS INTEGER,byval m as integer) as long dim Sum_Diag as long m=m-1 Sum_Diag=0 FOR i=0 to m step 1 Sum_Diag=Sum_Diag+A(i)(i) next i Diag=Sum_Diag end function private function InDiag(A()() AS INTEGER,byval m as integer) as long dim Sum_InDiag as long m=m-1 Sum_Diag=0 FOR i=0 to m step 1 Sum_InDiag=Sum_InDiag+A(i)(m-i) next i InDiag=Sum_InDiag end function二维数组靠边元素之和
