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Use trailing return type instead of deduction

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C++11 does not support deduction of return type.
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Lars Gullik Bjønnes 2018-02-15 17:34:38 +01:00 committed by Victor Zverovich
parent db86e8d5d3
commit f45f70af09
1 changed files with 2 additions and 2 deletions
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4
include/fmt/format.h
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@ -3199,10 +3199,10 @@ template <typename It, typename Char>
struct formatter<arg_join<It, Char>, Char>: struct formatter<arg_join<It, Char>, Char>:
formatter<typename std::iterator_traits<It>::value_type, Char> { formatter<typename std::iterator_traits<It>::value_type, Char> {
template <typename FormatContext> template <typename FormatContext>
auto format(const arg_join<It, Char> &value, FormatContext &ctx) { auto format(const arg_join<It, Char> &value, FormatContext &ctx) -> decltype(ctx.begin()) {
typedef formatter<typename std::iterator_traits<It>::value_type, Char> base; typedef formatter<typename std::iterator_traits<It>::value_type, Char> base;
auto it = value.begin; auto it = value.begin;
auto out = ctx.begin(); auto out = ctx.begin();
if (it != value.end) { if (it != value.end) {
out = base::format(*it++, ctx); out = base::format(*it++, ctx);
while (it != value.end) { while (it != value.end) {