I am trying to convert the first row of a long long matrix to long double, but when I print the result it only prints 0's. Here is the function:
我试图将长长矩阵的第一行转换为long double,但是当我打印结果时它只打印0。这是功能:
void convertLLtoLD(long long **B, long double **C, int n){
int i, j;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
C[i][j] = 0;
}
}
for (i = 0; i < n; i++) {
C[0][i] = B[0][i]; //C[0][i] = (long double)B[0][i];
}
for(i = 0; i < n; i++)
printf("%lf ", C[0][i]);
}
Here is the code where I allocate C that is done before calling the function:
这是我在调用函数之前分配C的代码:
long double** C = (long double**)malloc(n * sizeof(long double*));
for (i = 0; i < DIM; i++) {
C[i] = (long double*)malloc(n * sizeof(long double));
}
I'm compiling the code with GCC 4.8. B is read from a file and it's values are correct since I print them before I call the function. Even when I print them inside the function the result is the same. The result is also the same when I use a cast as in the comment. Can someone explain me why it isn't working?
我正在用GCC 4.8编译代码。从文件中读取B并且它的值是正确的,因为我在调用函数之前打印它们。即使我在函数内部打印它们,结果也是一样的。当我在评论中使用强制转换时,结果也是一样的。有人可以解释我为什么不工作?
Thanks
1 个解决方案
#1
5
In the statement :
在声明中:
printf("%lf ", C[0][i]);
You use an incorrect specifier (it expects a double
).
您使用不正确的说明符(它需要一个双精度数)。
Use %Lf
to print a long double
, use%lld
to print a long long
.
使用%Lf打印long double,使用%lld打印long long。
Note:
- In C++, just dont use
printf
, and preferstd::cout
and the standard library printf
specifiers from en.cppreference.com
在C ++中,只是不使用printf,而更喜欢std :: cout和标准库
en.cppreference.com的printf说明符
#1
5
In the statement :
在声明中:
printf("%lf ", C[0][i]);
You use an incorrect specifier (it expects a double
).
您使用不正确的说明符(它需要一个双精度数)。
Use %Lf
to print a long double
, use%lld
to print a long long
.
使用%Lf打印long double,使用%lld打印long long。
Note:
- In C++, just dont use
printf
, and preferstd::cout
and the standard library printf
specifiers from en.cppreference.com
在C ++中,只是不使用printf,而更喜欢std :: cout和标准库
en.cppreference.com的printf说明符