void inverse(char *p) {
if( *p = = '\\0' ) return;
inverse( p+1 ); printf( \*p ); }
int main(int argc, char *argv[]) {
inverse(\
return 0; }
借签了楼上的“递规反向输出” #include
void test(FILE *fread, FILE *fwrite) {
char buf[1024] = {0};
if (!fgets(buf, sizeof(buf), fread)) return;
test( fread, fwrite ); fputs(buf, fwrite); }
int main(int argc, char *argv[]) {
FILE *fr = NULL; FILE *fw = NULL; fr = fopen(\\ fw = fopen(\\ test(fr, fw); fclose(fr); fclose(fw); return 0; }
在对齐为4的情况下 struct BBB {
long num； char *name; short int data;
char ha; short ba[5]; }*p;
p=0x1000000; p+0x200=____; (Ulong)p+0x200=____;
(char*)p+0x200=____;
希望各位达人给出答案和原因，谢谢拉 解答：假设在32位CPU上， sizeof(long) = 4 bytes sizeof(char *) = 4 bytes
sizeof(short int) = sizeof(short) = 2 bytes sizeof(char) = 1 bytes
由于是4字节对齐，
sizeof(struct BBB) = sizeof(*p)
= 4 + 4 + 2 + 1 + 1/*补齐*/ + 2*5 + 2/*补齐*/ = 24 bytes (经Dev-C++验证)
p=0x1000000; p+0x200=____;
= 0x1000000 + 0x200*24
(Ulong)p+0x200=____; = 0x1000000 + 0x200
(char*)p+0x200=____; = 0x1000000 + 0x200*4
你可以参考一下指针运算的细节
写一段程序，找出数组中第k大小的数，输出数所在的位置。例如{2，4，3，4，7}中，第一大的数是7，位置在4。第二大、第三大的数都是4，位置在1、3随便输出哪一个均可。函数接口为：int find_orderk(const int* narry,const int n,const int k) 要求算法复杂度不能是O(n^2） 谢谢！
可以先用快速排序进行排序，其中用另外一个进行地址查找 代码如下，在VC++6.0运行通过。给分吧^-^
//快速排序
#include
usingnamespacestd;
intPartition (int*L,intlow,int high) {
inttemp = L[low]; intpt = L[low];
while (low < high) {
while (low < high && L[high] >= pt) --high;
L[low] = L[high];
while (low < high && L[low] <= pt) ++low;
L[low] = temp; }
L[low] = temp;
returnlow; }
voidQSort (int*L,intlow,int high) {
if (low < high) {
intpl = Partition (L,low,high);
QSort (L,low,pl - 1); QSort (L,pl + 1,high); } }
intmain () {
intnarry[100],addr[100]; intsum = 1,t;
cout << \number:\<< endl; cin >> t;
while (t != -1) {
narry[sum] = t;
addr[sum - 1] = t; sum++;
cin >> t; }
sum -= 1; QSort (narry,1,sum);
for (int i = 1; i <= sum;i++) cout << narry[i] << '\\t'; cout << endl;
intk;
cout << \input place you want:\<< endl; cin >> k;
intaa = 1; intkk = 0; for (;;) {
if (aa == k) break;
if (narry[kk] != narry[kk + 1]) {
aa += 1; kk++; }
}
cout << \NO.\<< k << \is:\<< narry[sum - kk] << endl;
cout << \it's place is:\; for (i = 0;i < sum;i++) {
if (addr[i] == narry[sum - kk]) cout << i << '\\t'; }
return0; }
1、找错 Void test1() {
char string[10];
char* str1=\23456789\
strcpy(string, str1);// 溢出，应该包括一个存放'\\0'的字符string[11] }
Void test2() {
char string[10], str1[10]; for(I=0; I<10;I++) {
str1[i] ='a'; }
strcpy(string, str1);// I，i没有声明。 }
Void test3(char* str1) {
char string[10];
if(strlen(str1)<=10)// 改成<10,字符溢出，将strlen改为sizeof也可以 {
strcpy(string, str1); } }
2. void g(int**); int main() {
int line[10],i;
int *p=line; //p是地址的地址 for (i=0;i<10;i++) {
*p=i;
g(&p);//数组对应的值加1 }
for(i=0;i<10;i++) printf(\return 0; }