C语言，在线等解答，明天考试

[陈思有]

#include <stdio.h>
#include <string.h>
void main()
{
	char (*name)[20] = {"peal","apple","strwaberll","cdfk","ndklfj"};
	int i,j,k;
	char *temp;

	for ( i=1; i<5; i++ )
	{
		temp = name[i];
		for ( j=i-1; j>=0; j-- );
		{
			if ( strcmp( temp ,name[j] ) < 1 )
				strcpy( name[j+1],name[j]);
			else
				break;
		}
		strcpy( name[j+1],temp);
	}

	for ( i=0; i<5; i++ )
	{
		printf("%s\n",name[i]);
	}
}

error C2078: too many initializers

[desdouble]

char (*name)[20] = {"peal","apple","strwaberll","cdfk","ndklfj"};

改为：

char * name[20] = {"peal","apple","strwaberll","cdfk","ndklfj"};

[科技完美生活]

strcpy的地方也有问题

[至善者善之敌]

char (*name)[20]
这个是数组指针
代表每个数组应该有20个元素
所以 char (*name)[20] = {"peal","apple","strwaberll","cdfk","ndklfj"};这里就有错误了

改成char *name[20]

另外下面的程序也有错，你单步调试吧

[desdouble]

#include <stdio.h>
#include <string.h>
int main()
{
char name[20][128] = {"peal","apple","strwaberll","cdfk","ndklfj"};
int i,j;
char temp[128] = {0};;

for (i = 0; i < 5; i++)
{
strcpy(temp, name[i]);
for (j = i + 1; j < 5; j++)
{
if ( strcmp( name[i],name[j] ) > 0 ) {
strcpy(temp, name[i]);
strcpy(name[i], name[j]);
strcpy(name[j], temp);
}
}
}
for ( i=0; i<5; i++ )
{
printf("%s\n",name[i]);
}

return 0;
}

排序结果：
double@ubuntu:~/Desktop$ ./tt
apple
cdfk
ndklfj
peal
strwaberll
double@ubuntu:~/Desktop$

[desdouble]

1、for ( j=i-1; j>=0; j-- );后面不应该有分号；
2、静态变量不能被写。

[hedy007]

char (*name)[20]
name是一个指针，指向char[20]
char *name[20]
name是一个数组，20个元素，均为指向char类型的指针。

[陈思有]

#include <stdio.h>
#include <string.h>
void main()
{
	char namep[5][15]={"peal","apple","strwaber","cdfk","ndklfj"};
	char (*name)[15]=namep;
	int i,j;
	char t[20],*temp=t;

	for ( i=1; i<5; i++ )
	{
		temp = name+i;
		for ( j=i-1; j>=0; j-- )
		{
			if ( strcmp( temp ,name+j ) < 0 )
				strcpy( name+j+1,name+j);
			else
				break;
		}
		strcpy( name+j+1,temp);
	}

	for ( i=0; i<5; i++ )
	{
		printf("%s\n",name+i);
	}
}

这是插入排序，输出的时候有些字符串被覆盖了，为什么

[标哥-iOS攻城狮]

#include <stdio.h>
#include <string.h>

void main()
{
//声明指针数组不能是(*name)[20],因为[]的优先级比*高，name先与[]结合，表示是一个数组，
//然后这个数组再与*结合，表示是一个指针数组
char name[][20] = {"peal","apple","strwaberll","cdfk","ndklfj"};
char (*pName)[20] = name; //定义一个行指针
int i,j, k;
char temp[20]; //这里定义成字符数组，如果是char *temp,它没有指向有大小的内存，
//就无法用于进行交换

for (i = 0; i < 5-1; i++ )
{
k = i;
for (j = i+1; j < 5; j++) //;去掉逗号
{
if (strcmp(pName[k], pName[j]) > 0 ) //不要用1，因为不一定是1
k = j;
}
if (k != i)
{
//交换需要一个中间变量
strcpy(temp, pName[k]);
strcpy(pName[k], pName[i]);
strcpy(pName[i], temp);
}
}

printf("字符串从小到大排列为： \n");
for (i = 0; i < 5; i++ )
{
printf("%s\n", pName[i]);
}
system("pause");
}

[dingshaofengbinbin]

[Quote=引用 1 楼 desdouble 的回复:]
char (*name)[20] = {"peal","apple","strwaberll","cdfk","ndklfj"};

改为：

char * name[20] = {"peal","apple","strwaberll","cdfk","ndklfj"};
[/Quote]
正解！！