英文:
Is it possible, in C, to write in one line a call to a function that has an array of strings (ie ptr) or int, or ... as parameter?
问题
无法在不进行变量声明的情况下直接调用函数 my_foo1 或 my_foo2。函数 my_foo1 需要一个字符指针的指针作为参数，而函数 my_foo2 需要一个整型指针作为参数。要调用这些函数，需要提供相应类型的变量或值作为参数，因此无法避免变量声明。
英文:
Let's consider these two functions :
void my_foo1(char ** my_par, int size) {
for (int i=0; i<size; i++) printf("%s \n",my_par[i]);
}
void my_foo2(int * my_par, int size) {
for (int i=0; i<size; i++) printf("%d \n",my_par[i]);
}
To call them, variables are declared and initialized. And after, function are called on a second line with these variables.
char * (my_strs[3])={"hello","world","!!!!"};
my_foo1(my_strs,3);
int my_ints[3]={1,2,3};
my_foo2(my_ints,3);
Is it possible to write something like :
my_foox(????,3)
and avoid the variable declaration ?
答案1
得分: 10
看起来你正在寻找的是复合文字：
my_foo1((char *[]){"hello","world","!!!!"},3);
my_foo2((int []){1,2,3},3);
请注意，这种文字的生命周期与其包围的范围相同。
英文:
It seems like what you're looking for is a compound literal:
my_foo1((char *[]){"hello","world","!!!!"},3);
my_foo2((int []){1,2,3},3);
Note that such literals have the lifetime of their enclosing scope.
答案2
得分: 2
你可以使用复合文字示例：
void my_foo2(int *my_par, int size) {
for (int i = 0; i < size; i++) printf("%d \n", my_par[i]);
}
my_foo2((int[]){1, 2, 3}, 3);
在函数foo2的调用中，具有类型int[3]的复合文字被隐式转换为其第一个元素的类型为int *的指针。
请注意，由于函数不会更改传递的数组，因此最好像这样声明它：
void my_foo2(const int *my_par, int size) {
for (int i = 0; i < size; i++) printf("%d \n", my_par[i]);
}
英文:
You can use compound literals as for example
void my_foo2(int * my_par, int size) {
for (int i=0; i<size; i++) printf("%d \n",my_par[i]);
}
my_foo2( ( int [] ){ 1, 2, 3 }, 3 );
In the call of the function foo2 the compound literal having the type int[3] is implicitly converted to a pointer of the type int * to its first element.
Pay attention to as the function does not change the passed array then it will be better to declare it like
void my_foo2( const int * my_par, int size) {
for (int i=0; i<size; i++) printf("%d \n",my_par[i]);
}
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