Question

我有一个数组：

CHAR m_manuf[256];

我正在尝试将值复制到此数组：

m_manuf = "abacus";  //This shows error

我也尝试过这种变化：

char abc[256] ="abacus";
m_manuf = abc; //Shows error as left value must be l-value

Answer 1

你不能＆＃39;像这样复制一个数组，而不是你能做到的，

operator_plus.hpp:164:1: note: template<class T1, class T2> typename arma::enable_if2<((arma::is_arma_sparse_type<T1>::value && arma::is_arma_sparse_type<T2>::value) && arma::is_same_type<typename T1::elem_type, typename T2::elem_type>::value), arma::SpGlue<T1, T2, arma::spglue_plus> >::result arma::operator+(const T1&, const T2&)
operator_plus.hpp:164:1: note:   template argument deduction/substitution failed:
operator_plus.hpp:164:1: error: no type named ‘result’ in ‘struct arma::enable_if2<false, arma::SpGlue<arma::SpMat<double>, arma::SpMat<std::complex<double> >, arma::spglue_plus> >’

或者

CHAR    m_manuf[256];
strcpy(m_manuf,"abacus" );

或者

char * m_manuf = "abacus";

注意：处理char数组的更好方法是使用std :: string，

Answer 2

一个常量的char数组对你来说已经足够了，所以你可以选择，

char *new = &tmp[0];

或者您需要修改新的char数组并且常量不正常，那么只需使用此

<?
// Your test array 1
$arr1[] = array('Title'=>'Ground','Price'=>'3.15');
$arr1[] = array('Title'=>'2 Day','Price'=>'12.11');
$arr1[] = array('Title'=>'1 Day','Price'=>'29.26');

foreach ($arr1 as $key => $value) {
    $newArr[$value['Title']] = $value['Price'];    
}

echo "<pre>";
print_r($newArr); // result of array1, you can follow this for other arrays as well
?>

C ++ char数组：复制数据时出错

2 个答案: