字符串数组问题。

[wjoin]

我想定义一个二维字符串数组，并把它传给另一个函数打印出来。
数组例：{“ddd","dfasdf","sadf","wer","ert"}
{"wer","ui,"twet","W","qwer"}
{"dfg","dr","23$","3wr"}

[kinglonghr]

上次有人问过这个问题
你用
char *[][];

[kinglonghr]

char *a[][4] ={
{"1","2"},
{"1","2","3"},
{"1","2"},
{"1","2","3","4"}
};

[晨星]

#include <stdio.h>

void printStrings(char* s[] , int n);

int main() {
char* s1[] = {"ddd","dfasdf","sadf","wer","ert"};
char* s2[] = {"wer","ui","twet","W","qwer"};
char* s3[] = {"dfg","dr","23$","3wr"};

printStrings(s1 , sizeof(s1) / sizeof(char *));
printStrings(s2 , sizeof(s2) / sizeof(char *));
printStrings(s3 , sizeof(s3) / sizeof(char *));

return 0;
}

void printStrings(char *s[] , int n) {
for(int i = 0 ; i < n ; i++)
printf("%s\n" , s[i]);
putchar('\n');
}

[wjoin]

怎么输出

[kinglonghr]

你去运行一下就知道了
printStrings函数执行这个输出功能

[wjoin]

steedhorse(晨星)
我是想不用：
printStrings(s1 , sizeof(s1) / sizeof(char *));
printStrings(s2 , sizeof(s2) / sizeof(char *));
printStrings(s3 , sizeof(s3) / sizeof(char *));
而只用调用一次printStrings()
就打印出所有内容

[liushuaiboy]

楼主的二维数组不是等长的，缺少的默认为0吗？二维数组可以简化成一维数组处理呀

[wjoin]

我是说把二维数组作为一参数传给一个打印函数，让这个打印函数打印出每个字符串

[liushuaiboy]

如果知道二维字符串数组的大小，当作参数传进去，按照steedhorse(晨星)的方法稍微改一下，就应该可以了，不会很难吧？^_^

[leonchew]

#include <stdio.h>

void printStrings(char* s1[][5] , int n);

int main() {
char* s[][5] = { {"ddd","dfasdf","sadf","wer","ert"},
{"wer","ui","twet","W","qwer"},
{"dfg","dr","23$","3wr"} };

printStrings( s , sizeof(s) / sizeof(s[0]) );

return 0;
}

void printStrings(char *s1[][5] , int n) {
int i,j;
for( i = 0; i < n; i++ ) {
for( j = 0; j < 5; j++ ) {
if( s1[i][j]!= NULL )
printf( " %s " , s1[i][j] );
}
putchar('\n');
}
}

自己运行试试吧。：）